<s>[INST] 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!
 [/INST] (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.
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