Solve in exact form: $x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$ ( WolframAlpha failed) 
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:

*

*I tried the Rational Root Thereom, but there is no rational root.

*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!
 A: (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.
A: 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.
A: 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.
A: 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.
