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(C) Given,$x$ is a cube root of unity other than $1$,so $x = \omega$ or $x = \omega^2$. Since $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$,we have $

Vedclass · Accepted Answer

$24$

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(C) Given,$x$ is a cube root of unity other than $1$,so $x = \omega$ or $x = \omega^2$. Since $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$,we have $\omega + \omega^2 = -1$. Consider the general term $T_n = \left(x^n + \frac{1}{x^n}ight)^2$. If $n$ is a multiple of $3$,$x^n = 1$,so $T_n = (1 + 1)^2 = 4$. There are $4$ such terms $(n=3, 6, 9, 12)$. If $n$ is not a multiple of $3$,$x^n$ is either $\omega$ or $\omega^2$,so $T_n = (\omega + \omega^2)^2 = (-1)^2 = 1$. There are $8$ such terms. The total sum is $8 	imes (1) + 4 	imes (4) = 8 + 16 = 24$.

If $x$ is a cube root of unity other than $1$,then $\left(x+\frac{1}{x}\right)^2+\left(x^2+\frac{1}{x^2}\right)^2+\ldots+\left(x^{12}+\frac{1}{x^{12}}\right)^2=$

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