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

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Solution

## Step 1: Find the roots of the equation :Let the roots of ${x}^{3}-2{x}^{2}+2x-1=0$ be $\alpha ,\beta ,\gamma$$⇒\left({x}^{3}-1\right)-1\left(2{x}^{2}-2x\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left({x}^{2}+x+1\right)-2x\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left({x}^{2}+x-2x+1\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left({x}^{2}-x+1\right)=0\phantom{\rule{0ex}{0ex}}⇒x-1=0⋮{x}^{2}-x+1=0$ Roots of ${x}^{3}-2{x}^{2}+2x-1=0$ are $1,\omega ,{\omega }^{2}$Step 2: Compute the required sum.${\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}={1}^{162}+{\left(\omega \right)}^{162}+{\left({\omega }^{2}\right)}^{162}\phantom{\rule{0ex}{0ex}}⇒{\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}=1+{\left({\omega }^{3}\right)}^{54}+{\left({\omega }^{3}\right)}^{108}\phantom{\rule{0ex}{0ex}}⇒{\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}=1+{\left(1\right)}^{54}+{\left(1\right)}^{108}\left[\because {\omega }^{3}=1\right]\phantom{\rule{0ex}{0ex}}⇒{\alpha }^{162}+{\beta }^{162}+{\gamma }^{162}=3$Hence, the sum of ${162}^{th}$ power of the roots of the equation ${x}^{3}-2{x}^{2}+2x-1=0$ is $3$.

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