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Solve the polynomial in closed form: $$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$$

WolframAlpha obviously failed.

I tried several ways:

  1. I tried the Rational Root Thereom, but there is no rational root.
  2. I tried possible factorisations

$$x^6-x^5+4x^4-4x^3+4x^2-x+1= (x^2+a_1x+a_2)(x^4+a_3x^3+a_4x^2+a_5x+a_6)$$

and

$$x^6-x^5+4x^4-4x^3+4x^2-x+1= (x^3+a_1x^2+a_2x+a_3)(x^3+a_4x+a_5)$$

But, the expansions are terrible!

user1094359
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    The polynomial is irreducible over $\mathbb Q[X]$. Since its galois-grroup over $\mathbb Q$ has $48$ elements (as checked with PARI/GP) , it can be solved by radicals but the expression can still be extremely complicated. I guess this is the case here. – Peter Sep 15 '22 at 11:16
  • @Peter Is it solvable group? – user1094359 Sep 15 '22 at 11:17
  • By applying Newton-Raphson method I found out that there is no solution $x\in\mathbb{R}$ – Vanessa Sep 15 '22 at 11:18
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    Since it has less than $60$ elements, it must be solvable. Aditionally , this follows from Burnsides theorem (only two prime factors). – Peter Sep 15 '22 at 11:18
  • @EveryonesaysFU This is actually true here, but I doubt we can prove this by using Newton's method. This method might fail even if there is a real solution. – Peter Sep 15 '22 at 11:20
  • @Peter Is my question bad? I get downvotes – user1094359 Sep 15 '22 at 11:20
  • Eureka! I solved it. – Vanessa Sep 15 '22 at 11:24
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    I played a little around with PARI/GP and found the following : If $s$ is the unique real solution of $s^3-s^2+s-2=0$ , then $x^2-sx+1$ is a factor of $f$. – Peter Sep 15 '22 at 11:27
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    @EveryonesaysFU. After such a comment, you must write what you did. – Claude Leibovici Sep 15 '22 at 11:34
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    You received just one downvote. I finally upvoted as well , but it would be nice to know where this polynomial came from. Perhaps in this case , it would also have been easier to spot the palindrome-property. Finally, you got the exact solution as desired. – Peter Sep 15 '22 at 12:03
  • @Peter Thank you! – user1094359 Sep 15 '22 at 12:12
  • A little bit more about palindromic polynomials in this answer of mine. I very much doubt that would be the first occurrence on our site. I recall using it in specific instances way earlier. Will delete the link, if a better/earlier can be located. – Jyrki Lahtonen Sep 19 '22 at 07:05

4 Answers4

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Your polynomial is a reciprocal polynomial, in that its coefficients read the same forwards to backwards. A consequence of this is that once you restrict to $x \neq 0$ (which is not a root) and divide by $x^3$, you get $$ x^3 + \frac 1{x^3} -\left(x^2+\frac 1{x^2}\right) + 4\left(x+\frac 1x\right) - 4 = 0 $$

Use the high school tricks given by $$ x^2+\frac 1{x^2} = \left(x+\frac 1x\right)^2 - 2 \\ x^3+\frac 1{x^3} = \left(x+\frac 1x\right)^3 - 3 \left(x+\frac 1x\right) $$

To get : if $z = x+\frac 1x$, then $$ (z^3-3z) - (z^2-2) + 4z - 4 = 0 \implies z^3-z^2+z-2 = 0 $$

Can this be solved using simple methods? The answer is no. This is a cubic polynomial : if it has to be reducible over the rationals, then it must have a linear factor. The rational root theorem fails, as one sees by trying the roots $\pm 1 , \pm 2$. Therefore, this polynomial is irreducible.

The only real root of this equation (that there is only one root can be found through calculus, for example) can be found explicitly through Cardano's method. It is given by $$ z = \frac 13\left(1 - 2 \sqrt[3]{\frac{2}{47+3 \sqrt{249}}} + \sqrt[3]{\frac 12(47 + 3\sqrt{249})}\right) $$

There are two other strictly complex roots $z_1$ and $\bar{z_1}$ (the complex conjugate of $z_1$) which drop out from this equation. They can be expressed in terms of $z$ as follows : $z_1\bar{z_1} = \frac 2{z}$ and $z_1+\bar{z_1} = 1-z$. This will tell you that $|z_1| = \sqrt{\frac{2}{z}}$ and $\text{Re}(z_1) = \frac{1-z}{2}$, so you can retrieve the complex part of $z_1$ and obtain an expression for $z_1$ entirely in terms of $z$.

Once you do this, you have three equations $$ x+\frac 1x = z , x+\frac 1x = z_1 , x+\frac 1x = \bar{z_1} $$

which can be solved as ordinary quadratic equations to yield $$ x = \frac{1}{2}\left(z\pm \sqrt{z^2-4}\right) \\ x = \frac{1}{2}\left(z_1\pm \sqrt{z_1^2-4}\right) \\ x = \frac{1}{2}\left(\bar{z_1}\pm \sqrt{\bar{z_1}^2-4}\right) $$

which are the six distinct roots of this problem. Note that all this detail is perhaps unimportant in computation : but it is useful if one is trying to compute relations between the roots, which is useful for computing Galois groups of these polynomials.

EDIT : A satisfying measure of how complicated it might be to find the roots of a polynomial is given by its Galois group. This edit partially addresses the concern that Wolfram was unable to factorize this polynomial as explicitly as I was, for example.

To provide an information explanation, the Galois group of a polynomial is a mathematical object that captures symmetries between roots. For example, consider the equation $x^3 - 1= 0$. The roots of this polynomial are given by $\omega , \omega^2$ and $\omega^3 = 1$ where $\omega$ is a complex cube root of unity. Notice the relations between the roots. This typically makes factoring easier.

A small Galois group (with respect to the biggest possible value of the size) typically indicates that a polynomial is "extremely simple". If you receive a high-degree polynomial which you're asked to factorize without a calculator (say, in an exam) then such a polynomial could have a very small Galois group (or is written very explicitly, typed wrongly etc. which makes the question easy for other reasons).

However, Galois groups are difficult to compute : and even fairly "obvious" symmetries such as reciprocity (also called Cohn or Gorenstein polynomials if some additional conditions are satisfied) are missed by engines that would much prefer numerical approximation to exact calculation if it meant a faster speed.

What follows is a vague explanation of how the Galois group is calculated in this case.

In this case, the most obvious symmetry is the relation that if $x$ is a root, so is $\frac 1x$. However, the reciprocal of the reciprocal returns you to the original number, so these relations are known as "involutions" because they are inverses of each other. There are three such involutions for three different roots, call them $r_1,r_2,r_3$.

Once one "forgets" about these involutions, then one is led to the polynomial $z^3-z^2+z-2$ that I derived earlier. This polynomial, as it turns out, doesn't have any polynomial relations among its roots. That has to do with a general result for cubic polynomials which is specified here.

From that thread, the Galois group of $z^3-z^2+z-2$ is of order $6$, the highest possible. In the case where the Galois group is of the highest possible order, the roots do not share any field relations with each other (so for example, even though $\bar{z_1}$ and $z_1$ are roots, it is not true that $\bar{z_1}$ is of the form $\frac{P(z_1)}{Q(z_1)}$ for $P,Q$ polynomials).

If these elements don't share any relations with each other, then you have $r_1,r_2,r_3$ which can be independently joined to any of these six permutations or not. That gives a choice of $6 \times 2^3 = \boxed{48}$ different permutations. The Galois group of $$ x^6-x^5+4x^4-4x^3+4x^2-x+1 $$ has size $48$ by making everything here rigorous.

The largest value that the Galois group could be? That is $6! = 720$.

Since $48$ is quite large but not very large, one is led to expect that there might be one or two simple relations between the roots, but not too many. This is reflected in what actually occurred.

Seeing that Galois group made me lick my lips. You've already seen why in the first part of this answer.

Sarvesh Ravichandran Iyer
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    Very well done! I might call it self-reciprocal or palindromic, but this is surely the simplest way forward. I was busily trying to figure out the Galois group, and failed to spot the symmetry. – Jyrki Lahtonen Sep 15 '22 at 11:51
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    I agree (+1). Does this also approve my above claim ? – Peter Sep 15 '22 at 11:53
  • @JyrkiLahtonen Thanks. I spent some time thinking about whether or not this was a duplicate of a general "reciprocal polynomial" technique, but when I realized that there might be more relations between the roots, I thought I'd just write an answer. – Sarvesh Ravichandran Iyer Sep 15 '22 at 11:53
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    @Peter Indeed, it does approve that conjecture. – Sarvesh Ravichandran Iyer Sep 15 '22 at 11:54
  • @Peter. Yes, this quickly leads to the conclusion that the Galois group is the wreath product $C_2\wr S_3$. Or the group of signed permutations of three coordinates, which is also the group of symmetries of a cube (rotations and reflections and combinations thereof). – Jyrki Lahtonen Sep 15 '22 at 11:56
  • This should be accepted. And by the way , I did not notice the palindrome property either :) – Peter Sep 15 '22 at 11:56
  • @Peter A more satisfying explanation is perhaps needed here than what I've given. There must be a reason why Wolfram couldn't solve this : that generally comes down to the relations between the roots being complex i.e. a fairly complex Galois group. For the sake of completeness, I'll see if I can make comments on the Galois-theoretic version of "why is this polynomial so complex?" Otherwise , I agree that it's fairly complete, but I'm not very satisfied with myself. – Sarvesh Ravichandran Iyer Sep 15 '22 at 12:06
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    I think , you can be proud of this work ! Even after having spot the palindromic property , the solution was far from obvious. – Peter Sep 15 '22 at 12:08
  • @Peter Thanks, yes I agree , I'm quite happy I noticed this. I'll go on to add the Galois section in any case to see if I can explain the Wolfram-Alpha failure somewhat. Thanks once again. – Sarvesh Ravichandran Iyer Sep 15 '22 at 12:09
  • https://math.stackexchange.com/q/4532181/1094359 If you want you can see – user1094359 Sep 15 '22 at 18:21
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    @user1094359 I was asleep when you posted this request, and got up to see two excellent answers, far more difficult than the one here. The Galois theory part helped there as well, because a group of order $72$, it's also in similar range to $48$ so you can expect at most one , perhaps two simplifications, but not too many. That's exactly what happened there , in fact Daniel went further into the Galois group logic in his answer. – Sarvesh Ravichandran Iyer Sep 16 '22 at 03:13
  • Thank you!!..Which answer do you think I should accept? I was undecided – user1094359 Sep 16 '22 at 10:54
  • @user1094359 The one that currently has $6$ votes is more basic, so probably more "pre-calculus" tailored. Personally, I would go for that one, but do overrule me if you wish to accept the other one. Both have a high number of votes, so accepting one would do a disservice on the other. That's unfortunate but , it is how it is! – Sarvesh Ravichandran Iyer Sep 16 '22 at 10:56
  • I also divided the polynomial by $x^2$.. But I couldn't see anything. The first answer found that is a perfect square I do not understand this. How did he/she observe this – user1094359 Sep 16 '22 at 11:09
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    @user1094359 I know that user for some time. Go through their answers, and you'll find that they've seen many problems like this. It comes with experience, but some users are genuinely talented. Also note that this user (in my knowledge) did not use any Galois group deductions (which the other author did, for example : how? I will find that slightly difficult to explain at a high school level) , so that makes their answer even more special. – Sarvesh Ravichandran Iyer Sep 16 '22 at 11:11
  • I had no doubt that you would also find that $P(x)/x^2$ contains a perfect square. I'm the only one who can't see.This could possibly be the original solution to the problem. Because it's totally precalculus. But I find Daniel's answer much more detailed. He also use precalculus. Although I don't understand what Galois is. I wish I could accept both. – user1094359 Sep 16 '22 at 11:34
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    @user1094359 Actually, I may not have had the idea of dividing by $x^2$ at all! If I had found the answer, it would have been more along the lines of Daniel. I agree that both answers are pre-calculus : but they're extremely difficult to ascertain, unlike this one which is quite standard. – Sarvesh Ravichandran Iyer Sep 16 '22 at 11:45
  • Thank you very much for your nice thoughts!. – user1094359 Sep 16 '22 at 11:58
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    @user1094359 You're welcome, thanks for consulting me and good luck with your future questions! – Sarvesh Ravichandran Iyer Sep 16 '22 at 11:58
  • I know I am disturbing you... But, can I get a very little but important information from you, if it is possible...?Could an under-voted or no-voted answer be useless or incorrect? Because an answer I got here seems doubtful solution to me, because it was almost never upvoted. Actually, I didn't vote either. Do you think this method could be mathematically invalid? Thank you very much for your guidance. I just wanted to know if it was valid. – user1094359 Sep 25 '22 at 20:00
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    @user1094359 Sure, I'll take a look and get back shortly. – Sarvesh Ravichandran Iyer Sep 26 '22 at 05:05
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    @user1094359 Yes, it's a valid method. Once $x = 0$ is ruled out, then lone student lets $u = x - \frac 1x$ and shows that there is a good relation between $x$ and $u$, which allows us to solve for $u$ in terms of $x$, leading to the answer. It's not voted on because it's the answer "at the bottom" so people don't see it, but I would argue that it's actually the better answer, because it actually explains how you get the cubic equation properly unlike the other one. This happens all the time on MSE : the best answer is not often the best voted answer. – Sarvesh Ravichandran Iyer Sep 26 '22 at 05:10
  • Got it, thanks a lot!.. I think you are the golden heart of MSE. I'm definitely sure I'm not the only one who thinks so. Thank you very much again! – user1094359 Sep 26 '22 at 09:37
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    @user1094359 Hey, thanks a lot! I can let you know about some other good contributors as well : people that I'm sure write more informative and excellent answers than me. There's so much to read, frankly. I think I'm "MSE" educated, and for good measure. A general audience doesn't know $0.01%$ of the magic that can be found here. – Sarvesh Ravichandran Iyer Sep 26 '22 at 10:04
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    There is more to the cubic $z^3-z^2+z-2=0$ than meets the eye. Kindly see my answer/comment below. – Tito Piezas III Nov 15 '22 at 10:25
  • +1 there especially for the second part about Galois groups. – Paramanand Singh Nov 17 '22 at 14:59
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(Too long for a comment.)

While the other answers addressed how to express the sextic in radicals, I was also curious about what other contexts it may arise. When faced with an equation, one trick is to look at its discriminant $d$ since it is an important invariant and may give clues. The OP's original equation is

$$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1 = 0\tag1$$

But this is palindromic so its degree can be reduced in half (as was done by S.Iyer's answer) to the cubic (which looks awfully familiar)

$$z^3-z^2+z-2=0\tag2$$

The discriminant of the sextic is $d_6 = -2^6\cdot83^2$ and the cubic is $d_3=83$. This is the clue. We have the class number $h(-83) = 3.$ Thus, the j-function $j(\tau)$ with $\tau = \tfrac{1+\sqrt{-83}}{2}$ is an algebraic integer of degree $3$ and it may be possible to express $j(\tau)$ using the real root $z$ of the cubic. In fact, one can do so and we have the cube,

$$j\left(\tfrac{1+\sqrt{-83}}{2}\right)=-32^3\left(\frac{25z+20}{23z-31}\right)^3\tag3$$

Therefore,

$$\quad\quad e^{\pi\sqrt{83}} \approx 32^3\left(\frac{25z+20}{23z-31}\right)^3 + 743.999999926\dots$$

just like its more famous cousin,

$$e^{\pi\sqrt{163}}\;\approx\; 640320^3 + 743.9999999\dots\quad$$

So it is curious in what context the OP found the sextic in the first place.

P.S. I may have seen $(2)$ in a table of cubics with small coefficients relating to class invariants.

Tito Piezas III
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    I'm extremely happy that an expert like you took interest in this question, and had absolutely no clue about the appearance of our old friend $744$! To me, the design of these questions doesn't seem like a coincidence, which is why I tried to search for a source for these questions (haven't asked the author yet). Some of their other questions are also gems. Thank you once again, and +1. – Sarvesh Ravichandran Iyer Nov 15 '22 at 10:52
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    @SarveshRavichandranIyer Thank you also for the kind words. Haven't been active in MSE for about 3 years (the last question I answered was in Aug 2019), but it's good to be back. – Tito Piezas III Nov 16 '22 at 12:31
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    +1 and glad to know that you are back in action! – Paramanand Singh Nov 17 '22 at 14:59
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The trick here is that the coefficients are palindromic. Let $$f(x)=x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1,$$ and let $$g(x)=f(x)/x^3=(x^3+1/x^3) - (x^2+1/x^2) +4(x+1/x)-4.$$ We can rewrite this in terms of $w=x+1/x$. We note that $w^2=(x+1/x)^2=x^2+1/x^2+2$, and $w^3=x^3+1/x^3 + 3(x+1/x)$, and so $x^2+1/x^2=w^2-2$ and $x^3+1/x^3=w^3-3w$.Therefore

$$\begin{split}g(x)&=(w^3-3w)-(w^2-2)+4w-4 \\ &=w^3-w^2+w-2=0\end{split}$$

Unfortunately, the roots of this cubic aren't particularly nice (as far as I can tell), but one can solve for them using the cubic formula, call them $r_1, r_2, r_3$, and then solve each of the equations of the form $x+1/x=r_i$, which yields a quadratic $x^2-r_i x +1=0$.

We remark that if the middle coefficient had been 5 instead of 4, then the resulting polynomial in w would have factored nicely and the problem would have a pretty solution.

Aaron
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The problem can be rewritten as $$x((x-1)x+1)(x^3+3x-1)+1=0$$ Now solving this we get $6$ roots:

$1.$ $-0.1399-1.7659i$

$2.$ $-0.1399+1.7659i$

$3.$ $-0.04270-0.56305i$

$4.$ $-0.04270+0.56305i$

$5.$ $0.67660-0.73635i$

$6.$ $0.67660+0.73635i$

These are all approximations I got from a software. Strict factorisation $($writing expression only in terms of factors$)$ is not really possible in my opinion or you can use these approx roots for that.

Vanessa
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