Denesting Phi, Denesting Cube Roots

Question

I have been looking into denesting square roots but I have found that $\sqrt[3]{2+\sqrt{5}}$ equals $(1+\sqrt{5})/2$. The same is true for $\sqrt[3]{2-\sqrt{5}}$ and $(1-\sqrt{5})/2$. I cannot figure how this is true. I proved this by setting both equal to x and forming polynomial equations. $y= x^2-x-1$ and $y=x^6-4x^3-1$. I do not know how to solve a degree 6 polynomial by hand like that. My objective is to denest the original $\sqrt[3]{2+\sqrt{5}}$ without knowledge of the simplified phi.

Answer 1

Hint : As a general rule, when dealing with nested radicals of the form $\sqrt[n]{A+B\sqrt[m]C}$ , you write $A+B\sqrt[m]C=(a+b\sqrt[m]C)^n$, and then employ Newton's binomial theorem. In our case, we have

$$\left(a+b\sqrt5\ \right)^3=a^3+3a^2(b\sqrt5)+3a(b\sqrt5)^2+(b\sqrt5)^3=\underbrace{(a^3+15ab^2)}_2+\underbrace{(3a^2b+5b^3)}_1\sqrt5$$ $$\iff a^3+15ab^2=2(3a^2b+5b^3)\iff a^3+15ab^2-6a^2b-10b^3=0\quad|:b^3\iff$$ $$\iff\left(\frac ab\right)^3-6\left(\frac ab\right)^2+15\left(\frac ab\right)-10=0\iff x^3-6x^2+15x-10=0\iff x=1$$ $$\iff\frac ab=1\iff a=b\iff a^3+15a^3=2\iff16a^3=2\iff a=\sqrt[3]\frac18=\frac12$$ or $3b^3+5b^3=1\iff b=\sqrt[3]\frac18=\frac12$ . Similarly for when the second term is $-1$.

Answer 2

Context for My Answer:

I came upon this question thanks to someone asking me on Discord, wanting to denest $$ \sqrt[4]{89 - 28 \sqrt{10}} \newcommand{\al}[1]{\left\{ \begin{align*} #1 \end{align*} \right.} \newcommand{\para}[1]{\left( #1 \right)} \newcommand{\pf}[2]{\para{ \frac{#1}{#2} }} \newcommand{\a}{\alpha} \newcommand{\b}{\beta} \newcommand{\g}{\gamma} $$ I found my way to Lucian's solution. However, I found it a bit difficult to parse through, especially when looking at it from the lens of a novice, so I felt a need to elaborate upon it and motivate it a bit more.

Throughout this post, then, I go through the process and details for denesting the above radical. (Ultimately the overall idea should be fairly similar to OP's examples, so having the additional, related, worked example should be helpful.)

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Overview/Preliminaries:

We seek to denest $$ \sqrt[4]{89 - 28 \sqrt{10}} \tag{1} $$ We consider the more general problem, $$ \sqrt[n]{A + B \sqrt[m]{C} } \tag{2} $$ $(1)$ is just $(2)$ with $(A,B,C,m,n)=(89,-28,10,2,4)$.

We will need the binomial theorem: recall, $$\begin{align*} &(x+y)^n = \sum_{k=0}^n \binom n k x^k y^{n-k} \\ &\text{ where } \binom n k = \frac{n!}{k!(n-k)!} \\ &\text{ and } k! = k(k-1)(k-2) \cdots (2)(1) \end{align*}$$

Now, returning to $(2)$, we want $a,b$ such that $$ A+B \sqrt[m]{C} = \para{ a+b \sqrt[m]{C} }^n $$ Well, we can expand the right-hand side by the binomial theorem. Notice that, then, $$ \para{ a+b \sqrt[m]{C} }^n = \sum_{k=0}^n \binom{n}{k} a^k \para{ b \sqrt[m]{C} }^{n-k} $$ From here, we'll note that $\a + \b \sqrt \g = \a' + \b' \sqrt{\g}$ if and only if $\a = \a'$ and $\b = \b'$. That is, we'll factor out our radical after expanding via the binomial theorem, set like coefficients equal, and do whatever manipulations necessary to get to an end result.

Broadly, these subsequent manipulations consist of the following:

• We'll have two equations, in two unknowns, our $a,b$.
• We'll multiply one equation by a constant, so the constant term in each is equal, so the variable terms must equal.
• Once we have an equation in integer coefficients, we divide by the highest power of $b$.
• The result is a polynomial in the variable $(a/b)$. We find its roots.
• If $r$ is a root, then this means $a/b=r \iff a=br$. We substitute this back in to our system of equations, to see if it results in solutions or not.

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Starting Out:

In $(1)$, then, we want $a,b$ such that $$ 89 - 28 \sqrt{10} = \para{ a + b \sqrt{10} }^4 $$ Well, expanding the right hand side, $$\begin{align*} \para{ a + b \sqrt{10} }^4 &=\binom 4 0 a^4 + \binom 4 1 a^3 b \sqrt{10} + \binom 4 2 a^2 \para{ b \sqrt{10} }^2 + \binom 4 3 a \para{ b \sqrt{10} }^3 + \binom 4 4 \para{ b \sqrt{10} }^4 \\ &= a^4 + 4 \sqrt{10} a^3 b + 60 a^2 b^2 + 40 \sqrt{10} a b^3 + 100 b^4 \end{align*}$$ We know $a,b$ need to be such that, $$ 89 - 28 \sqrt{10} = a^4 + 4 \sqrt{10} a^3 b + 60 a^2 b^2 + 40 \sqrt{10} a b^3 + 100 b^4 $$ so we factor on the right-hand side to get the parts with $\sqrt{10}$ and those without: $$ 89 - 28 \sqrt{10} = a^4 + 60 a^2 b^2 + 100 b^4 + \sqrt{10} \para{ 4 a^3 b + 40 a b^3 } $$ so we have the pair of equations $$\al{ 89 &= a^4 + 60 a^2 b^2 + 100 b^4 \\ -28 &= 4 a^3 b + 40 a b^3 } \tag{3}$$

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Dealing With Our System of Equations:

Now, look at the left-hand side of both equations. Clearly, if I multiply the second by $-89/28$, the equations are equal: $$ a^4 + 60 a^2 b^2 + 100 b^4 = - \frac{89}{7} a^3 b - \frac{890}{7} ab^3 $$ Let's make our lives just a tad easier by multiplying by $7$: $$ 7a^4 + 420 a^2 b^2 + 700 b^4 = -89a^3 b - 890 ab^3 $$ Move to the left-hand side: $$ 7a^4 + 89 a^3 b + 420 a^2 b^2 + 890 ab^3 + 700 b^4 = 0 $$ Divide through by $b^4$, our highest power of it. Then we get every term to have a power of $(a/b)$: $$ 7 \pf{a}{b}^4 + 89 \pf{a}{b}^3 + 420 \pf{a}{b}^2 + 890 \pf{a}{b} + 700 = 0 $$ This is a quartic equation in the variable $x = a/b$: $$ 7 x^4 + 89 x^3 + 420 x^2 + 890 x + 700 = 0 $$

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Utilizing the Roots:

Ultimately, you'll need to use whatever tools are at your disposal for solving polynomial equations. Inspection, graphing, the rational root theorem, it ultimately depends. Here, we can find that the roots of this are $$ x=-5 \qquad x=-2 \qquad x=- \frac{20 \pm 3i \sqrt{10}}{7} $$

We'll focus on the real roots, namely $x=-5$. Then $$ x = -5 \implies \frac a b = -5 \implies a = -5b $$ Consider our initial pair of equations in $(3)$. Replace $a$ with $-5b$ then. We get $$\al{ 89 &= (-5b)^4 + 60 (-5b)^2 b^2 + 100 b^4 \\ -28 &= 4 (-5b)^3 b + 40 (-5b) b^3 }$$ Simplifying, $$\al{ 89 &= 2225 b^4 \\ -28 &= -700 b^4 }$$ Either equation gives you $$ b^4 = \frac{1}{25} \implies b = \frac{1}{\sqrt{5}} $$ Then, since we know $a=-5b$, then $a = -5/\sqrt 5 = -\sqrt 5$.

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Conclusion:

This gives us $$ 89 - 28 \sqrt{10} = \para{ - \sqrt 5 + \frac{1}{\sqrt 5} \sqrt{10} }^4 = \para{ \sqrt 2 - \sqrt 5 }^4 $$ Taking fourth roots, then, we get $$ \sqrt[4]{89 - 28 \sqrt{10}} = \left| \sqrt 2 - \sqrt 5 \right| $$ (Recall that $\sqrt{x^2} = |x|$? The methodology is the same here.) Well, clearly, $$ \left| \sqrt 2 - \sqrt 5 \right| = \left| \sqrt 5 - \sqrt 2 \right| = \sqrt 5 - \sqrt 2 $$ since $\sqrt 5 > \sqrt 2$ and $|x-y|=|y-x|$ (just multiply by $-1$).

Finally, then, we have our denested expression: $$\bbox[lightblue,10px]{ \sqrt[4]{89 - 28 \sqrt{10}} = \sqrt 5 - \sqrt 2 } $$

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Addendum: What does the no-solution-from-a-root case look like?

Consider now a different case; what if $x=-1$ in our roots? Then $$ x = \frac a b = -1 \implies a = -b $$ Then our equations of concern in $(3)$ become $$\al{ 89 &= (-b)^4 + 60 (-b)^2 b^2 + 100 b^4 \\ -28 &= 4 (-b)^3 b + 40 (-b) b^3 }$$ and, with simplification, $$\al{ 89 &= 161 b^4 \\ -28 &= -44 b^4 }$$ Note that if you multiply the second equation by $-1$, and then by $89/28$ (so the left-hand sides would be equal), the coefficient on the right-hand side becomes $979/7 \approx 139.857 \ne 161$. In this case, then, no $b$ works -- so sometimes you'll have to try multiple different roots.