If x^{2}+x+1=0, then the value of (x+frac{1}{ | vedclass

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(D) Given $x^2+x+1=0$, the roots are the cube roots of unity, $\omega$ and $\omega^2$. Let $f(n) = (x^n + \frac{1}{x^n})^4$. Since $x^3=1$, $f(n)$ rep

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$145$

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(D) Given $x^2+x+1=0$, the roots are the cube roots of unity, $\omega$ and $\omega^2$. Let $f(n) = (x^n + \frac{1}{x^n})^4$. Since $x^3=1$, $f(n)$ repeats every $3$ terms. For $n=1$, $(x+\frac{1}{x})^4 = (\omega+\omega^2)^4 = (-1)^4 = 1$. For $n=2$, $(x^2+\frac{1}{x^2})^4 = (\omega^2+\omega)^4 = (-1)^4 = 1$. For $n=3$, $(x^3+\frac{1}{x^3})^4 = (1+1)^4 = 2^4 = 16$. The sequence of values is $1, 1, 16, 1, 1, 16, \dots$ There are $25$ terms in total. Number of full cycles of $(1, 1, 16)$ is $\lfloor 25/3 floor = 8$ cycles. Sum of $8$ cycles $= 8 	imes (1+1+16) = 8 	imes 18 = 144$. The $25^{th}$ term corresponds to $n=1$, which is $1$. Total sum $= 144 + 1 = 145$.

If $x^{2}+x+1=0$, then the value of $(x+\frac{1}{x})^{4}+(x^{2}+\frac{1}{x^{2}})^{4}+(x^{3}+\frac{1}{x^{3}})^{4}+\dots+(x^{25}+\frac{1}{x^{25}})^{4}$ is:

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