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!

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.
