The sum of the 162^{text{th}} power of the ro | vedclass

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(C) Given the equation $x^{3}-2x^{2}+2x-1=0$.By inspection,$x=1$ satisfies the equation as $1-2+2-1=0$.Thus,$(x-1)$ is a factor of the polynomial.Perf

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$3$

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(C) Given the equation $x^{3}-2x^{2}+2x-1=0$.By inspection,$x=1$ satisfies the equation as $1-2+2-1=0$.Thus,$(x-1)$ is a factor of the polynomial.Performing division,we get $(x-1)(x^{2}-x+1)=0$.The roots are $x=1$ and the roots of $x^{2}-x+1=0$,which are $x = \frac{1 \pm i\sqrt{3}}{2}$.These roots are $-\omega^{2}$ and $-\omega$,where $\omega$ is the complex cube root of unity.The sum of the $162^{	ext{th}}$ powers is $S = (1)^{162} + (-\omega^{2})^{162} + (-\omega)^{162}$.$S = 1 + \omega^{324} + \omega^{162}$.Since $\omega^{3} = 1$,we have $\omega^{324} = (\omega^{3})^{108} = 1$ and $\omega^{162} = (\omega^{3})^{54} = 1$.Therefore,$S = 1 + 1 + 1 = 3$.

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

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