How to solve $x^4-8x^3+24x^2-32x+16=0$ How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$ 

As $x\ne0,$ dividing either sides by $x^2$ 
$$x^2+\left(\frac4x\right)^2-8\left(x+\frac4x\right)+24=0$$
Now as $\displaystyle x^2+\left(\frac4x\right)^2=\left(x+\frac4x\right)^2-2\cdot x\cdot\frac4x$
Setting $x+\dfrac4x=y,$ we get $\displaystyle y^2-8-8y+24=0\implies(y-4)^2=0\iff y=4$
So, we have $\displaystyle x+\frac4x=4\iff(x-2)^2=0$
