Question

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

Answer 1

Step 1: Find the roots of the equation :

Let the roots of $x^3-2x^2+2x-1=0$ be $\alpha ,\beta ,\gamma$

$$\begin{gathered} \Rightarrow \left(x^3-1\right)-1(2x^2-2x)=0 \\ \Rightarrow (x-1)\left(x^2+x+1\right)-2x(x-1)=0 \\ \Rightarrow \left(x-1\right)\left(x^2+x-2x+1\right)=0 \\ \Rightarrow (x-1)(x^2-x+1)=0 \\ \Rightarrow x-1=0⋮x^2-x+1=0 \end{gathered}$$

Roots of $x^3-2x^2+2x-1=0$ are $1,\omega ,{\omega }^2$

Step 2: Compute the required sum.

$$\begin{gathered} {\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}=1^{162}+{\left(\omega \right)}^{162}+{\left({\omega }^2\right)}^{162} \\ \Rightarrow {\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}=1+{\left({\omega }^3\right)}^{54}+{\left({\omega }^3\right)}^{108} \\ \Rightarrow {\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}=1+{\left(1\right)}^{54}+{\left(1\right)}^{108}\left[\because {\omega }^3=1\right] \\ \Rightarrow {\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}=3 \end{gathered}$$

Hence, the sum of ${162}^{th}$ power of the roots of the equation $x^3-2x^2+2x-1=0$ is $3$.