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This is a review problem for an introductory Galois theory course.

Rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$.

There could be many ways to do this, but it's implicit that we use field theory for this problem. I tried looking at the solutions to this question, this question and this question, but they haven't helped me so far.

What seems clear is to replace $x = \sqrt[3]{5}$ and then our expression becomes $\frac{1}{1+x-x^2}$.

Now, since $x^3-5$ is the minimal polynomial of $\sqrt[3]{5}$ in $\mathbb Q[x]$, I think I want to find the inverse of $1+x-x^2$ in the quotient ring $\mathbb Q[x]/(x^3-5)$, but I'm not sure how.

Any hints or suggestions would be appreciated.

MathFail
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pyridoxal_trigeminus
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    Use the Euclidean algorithm to find a way to express $1$ as a multiple of $1+x-x^2$ plus a multiple of $x^3-5$. – Arturo Magidin May 10 '23 at 03:38
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    Alternatively, you can multiply the numerator and denominator by $(1+\omega\sqrt[3]{5}-\omega^2\sqrt[3]{25})(1+\omega^2\sqrt[3]5-\omega\sqrt[3]{25}),$ where $\omega$ is a primitive cube root of $1.$ Using that $\omega+\omega^2=-1.$ But that is harder than the Euclidean algorithm technique. – Thomas Andrews May 10 '23 at 03:51
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    For yet another alternative, you could write $\frac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}=a+b\sqrt[3]{5}+c\sqrt[3]{25}$ then cross-multiply and identify coefficients to find rational $a,b,c$ that satisfy the equation. But the $\gcd$ hint in the first comment is the better way to go. – dxiv May 10 '23 at 04:38
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    @User You should clarify in the description of the bounty what you are looking for which is not already covered in the questions linked by the OP, the previous comments above, or this other question. – dxiv May 13 '23 at 00:12
  • @dxiv I wrote the answer and then I deleted. Does this question require Galois theory ? – lone student May 13 '23 at 00:34
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    @lonestudent It does not require Galois theory. There are several different ways to solve it, as mentioned in my previous comment. That's why it's not clear to me what "canonical answer is supposed to mean in the description of the bounty. – dxiv May 13 '23 at 00:46
  • I posted an efficient way to do it using the Euclidean algorithm (so you can compare to other methods). Usually this is the most efficient (general) way to compute inverses of algebraic numbers. – Bill Dubuque May 14 '23 at 06:38

9 Answers9

7

Set $z=\sqrt[3]{5}$ and solve with the help of $z^3=5$ \begin{align*} (az^2+bz+c)\cdot (-z^2+z+1)&=1\\[6pt] -az^4+(a-b)z^3+(a+b-c)z^2+(b+c)z+c&=1\\[6pt] (a+b-c)z^2+(b+c-5a)z+(5a-b5+c)&=1\\[6pt] c=a+b\; , \;b=2a\; , \;1=5a-10a+a+2a=-2a&\\[6pt] a=-0.5\, , \,b=-1\, , \,c=-1.5& \end{align*} to get $$ \dfrac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}=\dfrac{1}{-z^2+z+1}=-\dfrac{1}{2}z^2-z-\dfrac{3}{2}=-\dfrac{1}{2}\sqrt[3]{25}-\sqrt[3]{5}-\dfrac{3}{2} $$

Marius S.L.
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    This method of undetermined coef's is usually much more work by hand than using the extended Euclidean algorithm (which here uses only two simple poly divisions). That's true in this case, but may not be obvious since most of the work is omitted above, such as computing the product polynomial and solving the system of equations. – Bill Dubuque May 18 '23 at 18:31
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We use $\color{red}{\text{Extended Euclidean Algorithm}}$ (for polynomials), which is a general method to deal with this type of rationalization problems, without introducing any tricks.

Let $P=x^3-5, Q=-x^2+x+1$, our goal is to express the constant term as the combination of $P$ and $Q$, such as

$$1=a(x)P+b(x)Q\Longleftrightarrow \frac{1}{Q}=\frac{a(x)P}{Q}+b(x)$$

If we set $x=\sqrt[3]{5}$, then term $P=0$, we get

$$\frac{1}{Q}=b(x)$$

So the term $b(x)$ will be the answer to this problem.

$$===============\text{START}===============$$

Do long division for $\frac{P}{Q}$, and we get

$$P=(-1-x)Q+\color{red}{R}\tag{1}$$

where $\color{red}{R}=2x-4$, do long division for $\frac{Q}{\color{red}{R}}$, and we get

$$Q=\left(-\frac{1}2x-\frac{1}2\right)\color{red}{R}-1\tag{2}$$

Plug eq.(1) into eq.(2) to eliminate $\color{red}{R}$, we get

$$1=\left(-\frac{1}2x-\frac{1}2\right)P+\left(-\frac{1}2x^2-x-\frac{3}2\right)Q$$

Set $x=\sqrt[3]{5}$, then term $P=0$, we get

$$\boxed{\frac{1}Q=\frac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}=-\frac{1}2\sqrt[3]{25}-\sqrt[3]{5}-\frac{3}2}$$

MathFail
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My goal will be to simply derive expressions in the denominator in terms of $\thinspace x^{3n}\thinspace . $

Indeed, $\thinspace x=\sqrt [3]{5}\thinspace $ leads to the fraction $\thinspace \frac{1}{1+x-x^2}\thinspace $ or $\thinspace -\frac{1}{x^2-x-1}\thinspace .$

Then, by applying the formula $\thinspace a^3\pm b^3=\left(a\pm b\right)\left(a^2\mp ab+b^2\right)\thinspace $ twice, you have :

$$ \begin{align}-\frac{1}{\left(x^2-x+1\right)-2}&=-\frac {x+1}{x^3+1-2(x+1)}\\ &=-\frac {x+1}{\left(x^3-1\right)-2x}\\ &=-\tfrac {(x+1)\left(\left(x^3-1\right)^2+2x\left(x^3-1\right)+4x^2\right)}{\left(x^3-1\right)^3-8x^3} \thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

Putting $\thinspace x=\sqrt [3]{5}\thinspace $, then the original fraction becomes :

$$ \begin{align}\frac{1}{1+\sqrt[3]{5}-\sqrt[3]{25}}&=-\frac 12 \big(3+2\sqrt [3]{5}+\sqrt [3]{25}\big)\end{align} $$

which completes the answer .

In the context of Bill Dubuque's explanatory comments ( 1 and 2 ) , the following method has also been added as a minor addition to the answer .

Let $\thinspace P(x)\thinspace $ be a polynomial. We want to multiply the denominator by a polynomial $\thinspace P(x)\thinspace $ such that the denominator can only be written in terms of $\thinspace x^{3n}$ :

$$ \begin{align}\frac {1}{1+x-x^2}&=-\frac {P(x)}{P(x)(x^2-x-1)}\end{align} $$

The construction of the polynomial $P(x)$ can be done as follows :

Let $\thinspace x=z\thinspace $ be one of the roots of $\thinspace x^2-x-1\thinspace .$ This implies that, $\thinspace z^2=z+1\thinspace $ and this leads to the following :

$$ \begin{align}z^6&=z^3+1+3z(z+1)\\ &=z^3+1+3z^3\\ &=4z^3+1\end{align} $$

Then, by quickly applying synthetic division, we have :

$$ \begin{align}&P(z)(z^2-z-1)=z^6-4z^3-1\\ \implies &P(z)=\frac {z^6-4z^3-1}{z^2-z-1}\\ \implies &P(z)=z^4+z^3+2z^2-z+1\end{align} $$

Putting $\thinspace x=\sqrt [3]{5}\thinspace $ and $\thinspace x^3=5\thinspace $, then we reach the following conclusion :

$$ \begin{align}\frac {1}{1+x-x^2}&=-\frac {x\left(x^3\right)+x^3+2x^2-x+1}{x^6-4x^3-1}\\ &=-\frac {2x^2+4x+6}{4}\\ &=-\frac 12 \big(3+2\sqrt [3]{5}+\sqrt [3]{25}\big)\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$

lone student
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    Neat calculation shortcut (+1). In the general case $\dfrac{1}{a_0+a_1x+a_2x^2}$ you would multiply with $(a_2 x - a_1)$ at the first step, then the rest works similarly, which gives another answer to How to rationalize denominator? – dxiv May 13 '23 at 02:25
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    @dxiv Thank you for mention the related link and reviewing the answer. I was very confused by the Galois theory in the tags of the question. (also, I have no idea how rationalizing the denominator relates to Galois theory). I stated in the first version of my answer that the computation lacks Galois theory and frankly, I planned to delete the answer. But, based on your comment above, I left it as it is . – lone student May 13 '23 at 03:55
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    @dxiv This is a special case of Gauss's inversion algorithm, i.e. it's the same as my answer except instead of modding the prior remainder by the current one we instead always mod the modulus $p=x^3-5$ by the current one. It corresponds to computing $(p,f) = 1$ for prime $p$ by using only Euclidean descent steps of form $(p,g) = (p, p\bmod g)$. This answer is equivalent to the special case $f$ is quadratic (so descent takes at most $2$ steps). If there is interest I can give further details – Bill Dubuque May 15 '23 at 16:53
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    @dxiv I appended a note to my answer showing in detail how the above answer is a special case of Gauss's inversion algorithm. At lone: it would be much clearer if you reduce the fractions using $x^3\equiv 5$ immediately (vs. at end). Then your first reduced fraction is $,-(x+1)/(4-2x),$ and the final arises by scaling top & bottom by $(4^3-(2x)^3)/(4-2x) = 4^2+4(2x)+(2x)^2,,$ as explained in my answer. – Bill Dubuque May 15 '23 at 21:05
  • @BillDubuque The understandable answer at the high school level as well, thanks . (+1) – lone student May 16 '23 at 01:15
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    The method you mention of computing the norm by elimination (e.g. by resultant) is so infrequently explicitly presented here that I think it is worth adding here to help it get better exposure – Bill Dubuque May 18 '23 at 21:31
  • How can this answer be correct? You noted earlier that $x=\sqrt [3]{5}$, even though the roots of the equation $x^2-x-1$ are $x=\frac {1\pm \sqrt{5}}{2}$. This answer is very confusing. – user1178051 May 19 '23 at 19:57
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    I suspect the method of "applying the formula" will not be clear to many readers at first glance. This can be remedied by elaborating on how it is applied, e.g. as I do in the first paragraph of my answer. – Bill Dubuque May 19 '23 at 20:05
  • The last part of above answer seems very confusing. What is the justification for this part? Would you like to explain @BillDubuque ? – user1178051 May 19 '23 at 20:07
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    @user1178051 lone seeks $p(x)$ so $,\dfrac{1}{f(x)}= \dfrac{1}{f(x)},\dfrac{p(x)}{p(x)},$ in $,\Bbb Q(x),,$ with $,f(x) = 1+x-x^2,,$ has denominator $,f(x)p(x)=g(x^3),\ g \in \Bbb Q[x],,$ so we get a rational denominator $,c=g(5)\neq 0$ when we eval at $,x = \sqrt[3]5,,$ so $,f^{-1} = p/c,$ is the sought inverse. Equivalently, working $!\bmod f(x),,$ we seek $,g(x^3)\equiv 0,,$ which lone finds by cubing $,x^2\equiv x+1.\ \ $ – Bill Dubuque May 19 '23 at 21:09
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    But instead of working $\bmod f = x^2-x-1,$ he instead works with a (generic) root $,z,$ of $f(x)$, which is equivalent by $,\Bbb Q[x]/(f(x))\cong \Bbb Q[z],,$ by $,f,$ irreducible over $\Bbb Q,,$ like $,\Bbb R[x]/(x^2+1)\cong \Bbb R[i] = \Bbb C.,$ Since we proved true the polynomial equality $,f(x)p(x)= g(x^3),$ in $,\Bbb Q[x],$ it remains true when evaluated at any value, such as $,x = \sqrt[3]5.\ \ $ – Bill Dubuque May 19 '23 at 21:09
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Now, since $x^3-5$ is the minimal polynomial of $\sqrt[3]{5}$ in $\mathbb Q[x]$, I think I want to find the inverse of $1+x-x^2$ in the quotient ring $\mathbb Q[x]/(x^3-5)$, but I'm not sure how.

While the posted answers are all worthy, as well as those under the related How to rationalize denominator?, the following is an attempt to stay closer to the above formulation of the question.

Idea is to  (1)  determine a cubic equation that $1+x-x^2$ satisfies in $\mathbb Q[x]/(x^3-5)$, then  (2)  multiply it by $(1+x-x^2)^{-1}$ and calculate the inverse.

(1)   Let $\,g = 1 + x - x^2\,$, then: $$ \begin{align} &g^3 - 3 g^2 + 18 g + 4 = -(x^3 - 5) (x^3 - 3 x^2 + 3 x + 4) \tag{1} \\ \implies\quad\;\; &g^3 - 3 g^2 + 18 g + 4 \equiv 0 \pmod{x^3 - 5} \tag{2} \end{align} $$ Polynomial $(1)$ was not magically pulled out of a hat. A step-by-step derivation using only elementary means is given for a more general case in my answer here to the question Is $a+5^{1/3}b+5^{2/3}c$ a root of any cubic polynomial in $\mathbb{Q}$? (this one here corresponds to $\lambda = \mu = 0, \nu = 5, a=b=1, c=-1$ in my other post).

(2)   $x^3-5\,$ is irreducible over $\mathbb Q$, so it is coprime to all rational polynomials of degree $\lt 3$. It follows that $\,g=1+x-x^2\,$ is invertible in $\mathbb Q[x]/(x^3-5)$, then multiplying $(2)$ by $g^{-1}$ and solving for $g^{-1}\,$: $$ \begin{align} &&4g^{-1} &\equiv -g^2 + 3 g - 18 \\ &&&\equiv -(1+x-x^2)^2 + 3 (1+x-x^2) - 18 \\ &&&\equiv -x^4 + 2 x^3 - 2 x^2 + x - 16 \\ &&&\equiv -(x^3 - 5)(x - 2) -2 (x^2 + 2 x + 3) \\ &&&\equiv -2 (x^2 + 2 x + 3) \pmod{x^3 - 5} \\ \implies\;\; &&(1+x-x^2)^{-1}\;\; &\equiv - \frac{1}{2}(3 + 2x + x^2) \pmod{x^3 - 5} \end{align} $$

dxiv
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  • (+1). Can all fractions with a finite number of algebraic radicals of higher degree always be rationalized? I haven't done research on this question before. For instance $$\frac {1}{\sqrt [17]{7}-\sqrt [7]{5}+\sqrt 2+1}$$ ? – lone student May 14 '23 at 04:39
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    @lonestudent Yes, any algebraic expression in algebraic numbers is algebraic itself, and can therefore be inverted/rationalized using some of the same techniques. – dxiv May 14 '23 at 04:55
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    This method is also already presented in many prior answers, e.g. see "generally" here where I present it from a more general perspective. – Bill Dubuque May 14 '23 at 16:35
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    I like the other answers as well, but I'm accepting this answer since it is the method I originally had in mind. – pyridoxal_trigeminus May 14 '23 at 17:39
  • @pyridoxal_trigeminus If you accept an answer that does not use Galois theory then it's best to remove that requirement from the question. – Bill Dubuque May 14 '23 at 17:43
  • @BillDubuque Fair enough. I changed the wording from "Galois theory" to "field theory" which perhaps is a better description of the method intended by the instructor of my course. – pyridoxal_trigeminus May 14 '23 at 17:50
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    @pyridoxal That makes more sense. Note that you can do it more Galois theoretically by multiplying by conjugates to get the norm, which is essentially the same as using the constant coef of some poly having it as root (as above) - see my answer linked in above comment. But for hand computation the forward Ext. Euclidean Algorithm is almost always (much) simpler, so I recommend that you learn that if you need to do such computations (if anything is unclear I can elaborate). – Bill Dubuque May 14 '23 at 18:00
  • @BillDubuque Thanks for that link. The method is well known, quite obviously, it's just that I thought it was relevant to OP's framing of the question, but didn't find it on cursory searches here and on approach0. It's no coincidence that your answer is in the more general context of algebraic elements, since in the case of algebraic expressions in the root of one single polynomial the solution based on Bezout's identity (or Euclidean division) is the simpler and more direct one. – dxiv May 14 '23 at 19:09
  • @pyridoxal The question is now essentially a dupe (of many), e.g. see the dupe links I gave here a month ago (which has yet another dupe answer- like all here). Even as originally restricted to Galois based methods it's a dupe. – Bill Dubuque May 14 '23 at 19:47
  • Wish the downvoter had left a comment why. – dxiv May 15 '23 at 03:38
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    @dxiv There will probably be no comment on this vote. Unfortunately :/ – lone student May 15 '23 at 09:11
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    At dxiv (and @pyridoxal ) I didn't downvote, but likely the reason is that this method is much more work than using the extended Euclidean algorithm (e.g. the easy method in my answer took me less than a minute by hand). The amount of work above is also hidden somewhat by omitting the computation of a polynomial having $g$ as root. – Bill Dubuque May 15 '23 at 16:25
  • @dxiv I had wanted to write something another answer 3-4 days ago, but I had left it halfway through, now I've completed it. But, I think it might me probably still a duplicate of something, but I don't know and I am not in control of the subject. I wanted to ask you before posting the answer. How can I show you this answer without posting it here? Is there a way? Unfortunately, the number of comments section characters is not enough :/ – lone student May 18 '23 at 17:01
  • @lonestudent There is no direct messaging facility on SE, and that's by design AFAICT. You could post the answer here and wait for users' feedback, or if you are unsure about a certain step specifically you could post it as a solution-verification question. – dxiv May 18 '23 at 18:24
  • @lone If you mean - as above - that we can trivially compute an inverse of $x$ from any poly killing it, i.e. $, 0 = f(x) = xg(x)-c \Rightarrow x^{-1}\equiv g(x)/c\pmod{!f(x)},$ then that is already in the answers linked in my first comment above, e.g. here. It's just a special case of the method of inversion by the extended Euclidean algorithm - when it takes only $1$ step: dividing $,f\div x,$ yielding $, f = xg + f(0),$ the Bezout identity for $,\gcd(f,x).\ \ $ – Bill Dubuque May 18 '23 at 18:25
  • But - as above - usually computing such a killing poly for the element being inverted is more work than inverting it by the extended Euclidean algorithm. See the linked answer for much more. $\ \ $ – Bill Dubuque May 18 '23 at 18:25
  • Thank you for important points and clarifications, dxiv . I think I found a way to share the answer, albeit temporarily . I would like to note this for @BillDubuque as well . – lone student May 18 '23 at 19:34
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    @lone in your method (revision $785$ here) we can use elimination (e.g. resultants) to compute the sought multiple (norm), e.g. $,{\rm resultant}(1+x−x^2,, y−x^3)=−y^2+4y+1,$. Resultants are a common way of computing norms (= product of conjugates), e.g. see here $\ \ $ – Bill Dubuque May 18 '23 at 20:12
  • @BillDubuque Thank you very much for reviewing the answer . So, since this way is also a duplicate, then I will not add this way in my current answer or post this as another answer . – lone student May 18 '23 at 20:22
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    @lonestudent The way I see it, the new answer combines ideas from different approaches in a novel way, so it's not a direct duplicate in that sense. However, it also does not bring something fundamentally different or better than previous answers, including your own here. – dxiv May 18 '23 at 21:00
  • @dxiv Thank you very much for reviewing the answer . Yes and you are right. I will not post the answer here or add to the existing answer. – lone student May 18 '23 at 21:13
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Let $\,x = \sqrt[3]5,\,$ so $\,\color{#90f}{x^3 = 5},\,$ so $\,(\color{#c00}{x\!-\!a})(\underbrace{x^2\!+\!ax\!+\!a^2}_{\textstyle \color{#b70}{g_a}})=\!\color{#90f}{\overbrace{x^3}^{5}}\!\!-a^3 = \begin{cases}\ \ \ \color{#0aa}{6,\,\ a=-1}\\[.1em] \color{#b70}{-3,\,\ a\ =\ 2}\end{cases}$

${\rm so} \ \ \dfrac1{(\color{#0aa}{x^2\!-\!x\!+\!1})\!-\!2}\ {\dfrac{x\!+\!1}{\color{#0aa}{x\!+\!1}}} =\dfrac{x\!+\!1}{\color{#0aa}6\!-\!2(x\!+\!1)} =\dfrac{\color{0a0}{3\!+\!\color{#b70}{x\!-\!2}\!}}{-2\,(\color{#b70}{x\!-\!2})}\ \color{#b70}{\dfrac{g_2}{g_2}}= \dfrac{3g_2\!\color{#b70}{-\!3}}{-2(\color{#b70}{-3})}=\dfrac{g_2\!-\!1}{2}$

That's the essence of lone student's proof. We show below how it is related to the more general algorithm for modular inversion by the extended Euclidean algorithm (in particular, we show how the above method is a special case of Gauss's algorithm for modular inversion).

We compute $\,\color{#c00}{f\equiv (x^2\!-\!x\!-\!1)^{-1}}\pmod{x^3\!-5}\,$ via Forward Extended Euclidean Algorithm

fractionally: $\ \ \ \ \dfrac{0}{\color{#90f}{x^3\!-5}}\overset{\large\frown}\equiv\dfrac{1}{x^2\!-\!x\!-\!1}\overset{\large\frown}\equiv\,\dfrac{\!\!-x\!-\!1}{\color{#0a0}{2x\!-\!4}}\overset{\large\frown}\equiv\color{#c00}{\dfrac{x^2\!+\!2x\!+\!3}2}\ $ i.e. in equation form:

$\!\!\!\begin{array}{rrl}\bmod\ \color{90f}{x^3-5}\!:\qquad\ \ [\![1]\!] &\color{#90f}{(x^3\!-5)}f&\!\!\!\equiv\, 0\\ [\![2]\!]&\!\! (x^2\!-x\!-\!1)f&\!\!\!\equiv\, 1;\ \ \ \ \ \ \ \ \ \ \color{#90f}{x^3\!-\!5} = \smash[t]{\overbrace{\color{#0aa}{(x\!+\!1)}}^{\rm quotient}(x^2\!-x\!-\!1) +\!\! \overbrace{\color{#0a0}{2x\!-\!4}}^{\rm remainder}}\to\\ [\![1]\!]\!-\!\color{#0aa}{(x\!+\!1)}[\![2]\!] = [\![3]\!] &\color{#0a0}{(2x\!-\!4)}f&\!\!\!\equiv -x\!-\!1;\ \ \ \ \ 2(x^2\!-x\!-\!1) = \color{#b70}{(x\!+\!1)}\color{#0a0}{(2x\!-\!4)}+\color{#c00}2\,\to\\ 2[\![2]\!]\!-\!\color{#b70}{(x\!+\!1)}[\![3]\!] = [\![4]\!] &\color{#c00}{2\:\!f}&\!\!\!\equiv\, \color{#c00}{x^2\!+\!2x\!+\!3}\\[-1em] \rm\small quotients\qquad\!\qquad &\rm\small remainders\ \ \ \ \end{array}$

Or since $\,p = x^3\!-\!5\,$ is prime we can use Gauss's inversion algorithm, so the last step is instead

$\!\!\begin{array}{rrl}[\![1]\!]\!-\!\color{#0aa}{(x\!+\!1)}[\![2]\!] = [\![3]\!] &\ \ \ \ \ \color{#0a0}{(2x\!-\!4)}f&\!\!\!\equiv -x\!-\!1; \ \ \ \ 2(x^3\!-\!5) = \smash[b]{\color{#b70}{\underbrace{(x^2\!+\!2x\!+\!4)}_{\large q(x) \,=\, g_2}}}\color{#0a0}{(2x\!-\!4)}+\color{#c00}6\,\to\\ 2[\![1]\!]\ -\ \color{#b70}{q(x)}\:\![\![3]\!] = [\![4]\!] &\color{#c00}{6f}&\!\!\!\equiv\, \color{#c00}{3(x^2\!+\!2x\!+\!3)} \end{array}$

Gauss's Algorithm uses the remainder sequence $\,r_{n+1} = p\bmod r_n\,$ (vs. $\,r_{n+1} = r_{n-1}\bmod r_n\,$ in the Euclidean algorithm). Writing the above in equivalent fraction form, using $x^3\equiv 5,\,$ we get

$$f \equiv \dfrac{1}{x^2\!-\!x\!-\!1}\,\color{#0aa}{\dfrac{x\!+\!1}{x\!+\!1}}\equiv\dfrac{x\!+\!1}{4\!-\!2x} \equiv \dfrac{x\!+\!1}{4\!-\!2x}\:\!\color{#b70}{\dfrac{x^2\!+\!2x\!+\!4}{x^2\!+\!2x\!+\!4}}\equiv\dfrac{3(x^2\!+\!2x\!+\!3)}{6}\qquad $$

The method in lone student's answer is exactly equivalent to that above - as can be seen by reducing lone's fractions $\!\bmod x^3\!-5$ (i.e. substitute $x^3\equiv 5)$ at every step (vs. at end). Note that the final step in lone's answer, i.e. scaling the fraction by $\,4\:\!\color{#b70}{q(x)}=\color{#08f}{(4^3\!-\!(2x)^3)/(4\!-\!2x)}\,$ is just another method of computing the final quotient $\:\!\color{#b70}{q(x)}$ when the divisor is linear $= x-a,\,$ and when $p = x^n - c.\,$ Namely, by the Polynomial Remainder Theorem we have

$$ p(x) = (x\!-\!a) q(x) + p(a) \,\Rightarrow\, \color{#b70}{q(x)} = \dfrac{p(x)-p(a)}{x-a} \left[= \color{#08f}{\dfrac{x^n-a^n}{x-a}}\ \ {\rm if}\ \ p = x^n-c\,\right]$$

Thus lone's method is a slight rearrangement of Gauss's inversion algorithm that works in the special case of inverting a quadratic $\,f\not\equiv 0\,$ modulo a prime $\ p = x^n-c.\,$ But Gauss's algorithm works more generally for any degree $f\not\equiv 0\,$ (and the extended Euclidean algorithm works more generally for coprime $p,f$ in any Euclidean domain, i.e. $p$ need not be prime - only coprime to $f)$

Bill Dubuque
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  • Likely the simplest (general) way. $ $ OP is $,\color{#c00}{{-}f \equiv -\frac{1}2(x^2!+!2x!+!3)},\ x= \sqrt[3]{5}\ $ (posted only for comparison to Galois-theoretic methods) – Bill Dubuque May 14 '23 at 06:30
  • Equation Arithmetic: $,\color{#0a0}{ a[![j]!]} + \color{#c00}{b[![k]!]} =: [![\ell]!],$ means that equation number $,\ell,$ is obtained by scaling equation $\color{#0a0}j$ by $\color{#0a0}a$ then adding it to the scaling of equation $\color{#c00}k$ by $\color{#c00}b.,$ The sequence of coef's of $f$ are the remainders in the Euclidean algorithm for $\gcd(x^3!-!5,x^2!-!x!-!1),,$ as explained here (for integers) – Bill Dubuque May 14 '23 at 18:42
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Applying formulas for sum and difference of cubes, we can get immediately: $$\dfrac1{1+\sqrt[\large3]5-\sqrt[\large3]{25}} =\dfrac1{2-(1-\sqrt[\large3]5+\sqrt[\large3]{25})} =\dfrac{1+\sqrt[\large3]5}{2(1+\sqrt[\large3]5))-6} = \dfrac{\sqrt[\large3]5+1}{2(\sqrt[\large3]5-2)}$$ $$= \dfrac{(\sqrt[\large3]5+1)(\sqrt[\large3]{25}+2\sqrt[\large3]5+4)}{2(5-8)}=-\dfrac{9+6\sqrt[\large3]5+3\sqrt[\large3]{25}}6 =\color{green}{\mathbf{ -\dfrac12(\sqrt[\large3]{25}+2\sqrt[\large3]5+3)}.}$$

Evidently, is not a problem to express these calculations chain in the terms of the Galois's theory.

Yuri Negometyanov
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    This is the same as lone student's answer (if you reduce $x^3\to 5$ at every step (vs. end), as I suggested yesterday in comments there). My answer shows it's a special case of Gauss's modular inversion algorithm. – Bill Dubuque May 16 '23 at 17:07
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By polynomial long division, $$\frac{x^3-5}{-x^2+x+1}=-x-1+\frac{2x-4}{-x^2+x+1}=0\tag1$$ $$x\frac{x^3-5}{-x^2+x+1}=-x^2-x-2+\frac{-2x+2}{-x^2+x+1}=0\tag2$$ and by adding $(1)$ and $(2)$ we have $-x^2-2x-3-\frac{2}{-x^2+x+1}=0$ and $\frac{1}{-x^2+x+1}=-\frac12x^2-x-\frac32$.

Bob Dobbs
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    This is just the extended Euclidean algorithm for $,\gcd(x^3−5,f), f=−x^2+x+1,,$ expressed in fraction form (usually done in continued fraction form). It is exactly the same as the first Extended Euclidean algorithm computation in my answer, if you combine your last two steps by multiplying equation $(1)$ by $,\color{#c50}{x+1},,$ instead of $,x,,$ i.e. $(2)+(1)=\color{#c50}x(1)+\color{#c50}1(1)=(x+1)(1),,$ i.e. compute the quotient $−2(−x^2+x+1)\div\color{#0a0}{(2x−4)}=\color{#c50}{x+1},$ in one (vs. your two) steps, as in equation $[![4]!]$ in my answer. $\ \ $ – Bill Dubuque May 18 '23 at 10:16
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    Note: in my answer I compute the inverse of the negation $−f,,$ so signs differ. Also, your final result upon adding is not $=0$ but is $\equiv 0\pmod{!x^3-5},$ or, equivalently, it is $=0$ when evaluated at $,x = \sqrt[3]5.\ \ $ – Bill Dubuque May 18 '23 at 10:22
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Let $x = \sqrt[3]{5}$ and let $y = \frac{1}{1 + x - x^2}$ (the value you want to solve for). Then:

$$(1 + x - x^2)y = 1 \tag{1}$$

Multiplying by $x$ (and simplifying $x^3 = 5$) gives:

$$(x + x^2 - 5)y = x \tag{2}$$

Multiplying by $x$ again gives:

$$(x^2 + 5 - 5x)y = x^2 \tag{3}$$

And multiplying by $x$ one more time just takes us back to equation (1), except scaled by a factor of 5.

Let's consider a weighted combination of the three equations we have, for some constants $A$, $B$, and $C$.

$$A(x^2 + 5 - 5x) + B(x + x^2 - 5) + C(1 + x - x^2) = \frac{Ax^2 + Bx + C}{y}$$ $$(A + B - C)x^2 + (-5A+B+C)x + 5A - 5B + C = \frac{Ax^2 + Bx + C}{y}$$

Try to pick constants that make the left side as simple as possible. Let's eliminate the $x^2$ term by setting $C = A + B$.

$$(-4A+2B)x + 6A - 4B = \frac{Ax^2 + Bx + A + B}{y}$$

Now eliminate the $x$ term by setting $B = 2A$.

$$-2A = \frac{Ax^2 + 2Ax + 3A}{y}$$

And now the $A$'s just cancel out.

$$-2 = \frac{x^2 + 2x + 3}{y}$$ $$y = \frac{-x^2 - 2x - 3}{2}$$ $$\boxed{y = \frac{-\sqrt[3]{25} - 2\sqrt[3]{5} - 3}{2}} \approx -4.671985$$

Bill Dubuque
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Dan
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    This is equivalent to inverting $, \beta = 1+x-x^2,$ by computing its norm as the determinant of the multiplication by $\beta$ map (e.g. see here) or using resultants, as I mentioned a few hours ago in comments on dxiv's answer. – Bill Dubuque May 19 '23 at 02:36
  • Simpler: to compute the inverse $,y,$ we use elimination on your equations $(1),(2),(3)$ of form $, g_i, y = f_i,$ to decrease the degree of the $g_i$ (= denominator of $,y = f_i/g_i),,$ yielding

    $$\begin{align} \color{#0a0}{(4)}\ :=\ (1)+(2) &\ \ \ \ \ \ \ \ (2x-4), y =\ \ x+1\[.2em] \color{#90f}{(5)}\ :=\ (1)+(3)&\ \ \ \ \ (-4x+6),y = x^2+1\[.2em] (6):=2\color{#0a0}{(4)}+\color{#90f}{(5)}&\qquad\quad\ \ \ \ \ :!\color{#c00}{{-}2, y = x^2+2x+3}.\ \ \ \bf\small QED \end{align}$$

     – Bill Dubuque May 19 '23 at 19:13
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    This is very closely related to inversion by the extended Euclidean algorithm (cf. my answer) where we use Euclidean division (with remainder) to find smaller degree denominators. – Bill Dubuque May 19 '23 at 19:13
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Yet an other solution, that covers the general case: $$ \bbox[lightblue]{\qquad \xi=\xi(a) :=\frac 1{1+\sqrt[3]a-\sqrt[3]{a^2}} =\frac 1{1+A-A^2} \ , \qquad A:=\sqrt[3]a\ .\qquad } $$ The problem comes with $a=5$, the solution works in a more general case. But let us focus on this special $a=5$. Some motivation first. The number $\xi$ above lives in the field of algebraic numbers $\Bbb Q(\sqrt[3]{a}) =\Bbb Q(A)$. Which is not a Galois field. We may pass to a Galois closure $K$ of it, and then the norm $N_{K:\Bbb Q}(1+A-A^2)$ is:

  • a rational number of the one side,
  • a product of all conjugates of $\alpha_1:=1+\sqrt[3]{a}-\sqrt[3]{a^2} =1+A-A^2$ in $K$, which are $\alpha_1$, the same number, and some other numbers $\alpha_2,\alpha_3,\dots$

And from here, we already know what to do, we have $\xi=\frac 1{\alpha_1}=\frac {\phantom{\alpha_1}\alpha_2\alpha_3\dots}{\alpha_1\alpha_2\alpha_3\dots}=\frac{\alpha_2\alpha_3\dots}{N_{K:\Bbb Q}(\alpha_1)}$, and that norm in the denominator is a rational number.

Sometimes, it is enough to use only some of the conjugates, and here is so the case. We can start the computation / the solution:

Let $\varepsilon$ be a primitive root of order three of the unit, e.g. $\frac 12(-1+\sqrt{-3})$. We need only $\varepsilon^2+\varepsilon+1=0$. Consider the following numbers, obtained from $\alpha_1$ by (formally) replacing $\sqrt[3]{a}$ by itself, by $\varepsilon\sqrt[3]a$, and by $\varepsilon^2\sqrt[3]a$: $$ \begin{aligned} \alpha_1 &&&= 1+A-A^2 &&= 1+\sqrt[3]{a}-\sqrt[3]{a^2}\ ,\\ \alpha_2 &&&= 1+\varepsilon A-\varepsilon^2 A^2 && =1+\varepsilon\sqrt[3]{a}-\varepsilon^2\sqrt[3]{a^2}\ ,\\ \alpha_3 &&&= 1+\varepsilon^2 A-\varepsilon A^2 && =1+\varepsilon^2\sqrt[3]{a}-\varepsilon\sqrt[3]{a^2}\ . \end{aligned} $$ Recall the identity: $$ \begin{aligned} X^3 + Y^3 + Z^3 &-3XYZ \\ &=(X+Y+Z)(X^2 +Y^2+Z^2-YZ-ZX-XY) \\ &=(X+Y+Z)(X+\varepsilon Y+\varepsilon^2 Z)(X+\varepsilon^2 Y+\varepsilon Z) \ . \\[3mm] &\qquad\text{ Apply it for:}\\ X&=1\ ,\\ Y&=A=\sqrt[3]{a}\ ,\\ Z&=-A^2=-\sqrt[3]{a^2}\ .\\ &\qquad\text{ Then we get:}\\[3mm] \alpha_1\alpha_2\alpha_3 &=X^3+Y^3+Z^3-3XYZ=1+a-a^2+3a=1+4a-a^2\in\Bbb Q\ ,\\ \alpha_2\alpha_3 &=X^2+Y^2+Z^2-YZ-ZX-XY \\ &=1+A^2+A^4 +A^3+A^2-A\\ &=1+A^2+aA+a+A^2-A\\ &=(1+a)+(a-1)A+2A^2\ . \end{aligned} $$

For $\bbox[yellow]{\ a=5\ }$ we obtain: $$ \begin{aligned} \xi=\xi(5) &= \frac1{1+\sqrt[3]5-\sqrt[3]{25}} = \frac{\alpha_2\alpha_3}{\alpha_1\alpha_2\alpha_3} = \frac{6+4A+2A^2}{1+20-25} = -\frac 14(6+4A+2A^2) \\ &=-\frac 12(3+2A+A^2) =\bbox[yellow]{\ -\frac 12\left(\ 3 + 2\sqrt[3]5 + \sqrt[3]{25}\ \right)\ }\ . \end{aligned} $$

Computer check: Here using sage.

sage: R.<x> = PolynomialRing(QQ)
sage: K.<A> = NumberField(x^3 - 5)
sage: 1/(1 + A - A^2)
-1/2*A^2 - A - 3/2
dan_fulea
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  • Galois theoretical arguments have been already mentioned in one solution, and in many comments. Initially, as i saw this direction already mentioned, i was less attracted to double a solution. However, seeing the many other solutions - where the landscape becomes computationally really dirty, without being quick & dirty, i decided to post the above showing: $(1)$ what to do in the more general case, e.g. with $$\frac1{s+tA+uA^2}\ , \ A=\sqrt[3]a\ ,\ a,s,t,u\in\Bbb Z\ ,$$ and $(2)$ give the identity - which also solves the cubic, $X^3+Y^3+Z^3-3XYZ=\dots$ and is the core of the computation(s) – dan_fulea May 19 '23 at 13:29