Question

The sum of ${162}^{\text{th}}$ power of the roots of the equation $x^3-2x^2+2x-1=0$ is

Answer 1

Let roots of $x^3-2x^2+2x-1=0$ be $\alpha ,\beta ,\gamma .$
$(x^3-1)-(2x^2-2x)=0$
$\Rightarrow (x-1)(x^2-x+1)=0$
$\Rightarrow x=1,-\omega ,-{\omega }^2$

Now, ${\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}$
$=1+(-\omega {)}^{162}+(-{\omega }^2{)}^{162}$
$=1+({\omega }^3{)}^{54}+({\omega }^3{)}^{108}=3$