Question

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

• A. 12
• B. 64
• C. 24
• D. 0

Answer 1

Answer: (C)

Given, $x$ is a cube root of unity other than 1 i.e. $x=\omega$ or $\omega^{2}$
Now, $\left(x+\frac{1}{x}\right)^{2}+\left(x^{2}+\frac{1}{x^{2}}\right)^{2}+\ldots+\left(x^{12}+\frac{1}{x^{12}}\right)^{2}$
$=\left(\omega+\frac{1}{\omega}\right)^{2}+\left(\omega^{2}+\frac{1}{\omega^{2}}\right)^{2}+\ldots+\left(\omega^{12}+\frac{1}{\omega^{12}}\right)^{2}$
$=\left(\omega+\frac{1}{\omega}\right)^{2}+\left(\omega^{2}+\frac{1}{\omega^{2}}\right)^{2}+\ldots+\left(\omega^{11}+\frac{1}{\omega^{11}}\right)^{2}$
$=\left(\omega+\omega^{2}\right)^{2}+\left(\omega^{2}+\omega\right)^{2}+(1+1)^{2}+\left(\omega+\omega^{2}\right)^{2}$
$+\left(\omega^{2}+\omega\right)^{2}+(1+1)^{2}+\left(\omega+\omega^{2}\right)^{2}$
$=8\left(\omega+\omega^{2}\right)^{2}+4(1+1)^{2}$
$=8(-1)^{2}+4(2)^{2}=8+16=24$