Solving $x^4 -10x^3 + 26x^2 -10x +1 = 0$. Find the root of the  equation 
$$x^4 -10x^3 + 26x^2 -10x +1 = 0$$ 
My attempt:
If $a,b,c$ and $d$ are a roots then $a+b+c+d = 10,$
$(a+b)(c+d)  + ab +cd = 26$
$(a+b)cd  +ab(c+d) = 10$
$abcd =1$
Now  i'm not able  procceed further.
 A: You can proceede also like this:
\begin{eqnarray}x^4 -10x^3 + 26x^2 -10x +1 &=& (x^4 -4x^3 + 6x^2 -4x +1)-6x(x^2-2x+1)+8x^2\\
&=& (x-1)^4-6x(x-1)^2+8x^2\\
&=& \Big((x-1)^2-4x\Big)\Big((x-1)^2-2x\Big)\\
&=& \Big(x^2-6x+1\Big)\Big(x^2-4x+1\Big)\\
\end{eqnarray}
A: .For palindromic polynomials (whose coefficients read the same from left to right), it is key to realize that for $x \neq 0$, we have $f(x) = 0$ if and only if $f(\frac 1x) = 0$. This is a routine check which I leave to you.
More precisely, note that $x = 0$ is not a root of the polynomial above. Therefore, dividing by $x^2$ and collecting terms, we have :
$$
x^2 + \frac 1{x^2} - 10\left(x + \frac 1x\right) + 26 = 0
$$
Note that $x^2 + \frac 1{x^2} = \left(x + \frac 1x\right)^2 - 2$. Therefore, setting $x + \frac 1x = y$, we get a quadratic polynomial in $y$, which can be solved to get two values of $y$, followed by four values of $x$.

More precisely, $y^2 -10y + 24 = 0$.  So we get $y  = 4,6$ so $x + \frac 1x = 4$ or $x + \frac 1x  = 6$. Solve these to get $2 \pm \sqrt 3$ and $3 \pm 2 \sqrt 2$. Either from here, or from the earlier equations you may notice that they come in pairs whose product is $1$.
