Let A_r = left(x+frac{1}{x}right)^3 cdot left | vedclass

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(D) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$,where $\omega^3=1$ and $1+\omega+\omega^2=0$. Note that $\omega + \frac{1}{\omega} = \omeg

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$\frac{1}{7}$

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(D) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$,where $\omega^3=1$ and $1+\omega+\omega^2=0$. Note that $\omega + \frac{1}{\omega} = \omega + \omega^2 = -1$. Thus,the terms in the product $A_r$ are: For $k=1$,$(x+\frac{1}{x})^3 = (-1)^3 = -1$. For $k=2$,$(x^2+\frac{1}{x^2})^3 = (\omega^2+\omega)^3 = (-1)^3 = -1$. For $k=3$,$(x^3+\frac{1}{x^3})^3 = (1+1)^3 = 8$. So,$A_3 = (-1)(-1)(8) = 8$. For $k=4$,$(x^4+\frac{1}{x^4})^3 = (\omega+\omega^2)^3 = -1$. For $k=5$,$(x^5+\frac{1}{x^5})^3 = (\omega^2+\omega)^3 = -1$. For $k=6$,$(x^6+\frac{1}{x^6})^3 = (1+1)^3 = 8$. Thus,$A_6 = A_3 \cdot (-1) \cdot (-1) \cdot 8 = 8^2 = 64$. In general,$A_{3n} = 8^n$. The series is $\sum_{n=1}^{\infty} \frac{1}{A_{3n}} = \sum_{n=1}^{\infty} \frac{1}{8^n} = \frac{1/8}{1-1/8} = \frac{1/8}{7/8} = \frac{1}{7}$.

Let $A_r = \left(x+\frac{1}{x}\right)^3 \cdot \left(x^2+\frac{1}{x^2}\right)^3 \cdot \left(x^3+\frac{1}{x^3}\right)^3 \cdots \left(x^r+\frac{1}{x^r}\right)^3$. If $x^2+x+1=0$,then $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\cdots \infty =$

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