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.

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}
