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(C) The roots of the equation $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.Given $\alpha = \omega$ and $\bet

Vedclass · Accepted Answer

$24$

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(C) The roots of the equation $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.Given $\alpha = \omega$ and $\beta = \omega^2$,we have $\alpha+\beta = -1$ and $\alpha\beta = 1$.For any $n$,$\alpha^n+\beta^n = \omega^n+\omega^{2n}$.If $n$ is a multiple of $3$,then $\omega^n = 1$ and $\omega^{2n} = 1$,so $\alpha^n+\beta^n = 1+1 = 2$.If $n$ is not a multiple of $3$,then $\alpha^n+\beta^n = \omega^n+\omega^{2n} = -1$.We need to evaluate $S = \sum_{n=1}^{12} (\alpha^n+\beta^n)^2$.For $n=1, 2, 4, 5, 7, 8, 10, 11$ ($8$ terms),the value is $(-1)^2 = 1$.For $n=3, 6, 9, 12$ ($4$ terms),the value is $(2)^2 = 4$.Thus,$S = 8 	imes (1) + 4 	imes (4) = 8 + 16 = 24$.

If $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$,then $(\alpha+\beta)^2+(\alpha^2+\beta^2)^2+(\alpha^3+\beta^3)^2+\ldots+(\alpha^{12}+\beta^{12})^2=$

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