If \(x\) is a cube root of unity other than 1 , then…
If \(x\) is a cube root of unity other than 1 , then \(\left(x+\frac{1}{x}\right)^2+\left(x^2+\frac{1}{x^2}\right)^2+\ldots+\left(x^{12}+\frac{1}{x^{12}}\right)^2=\)
12
64
24
0
Solution
Given, \(x\) is a cube root of unity other than 1 i.e. \(x=\omega\) or \(\omega^2\)
Now
\(\begin{aligned}
& \left(\quad\left(x+\frac{1}{x}\right)^2+\left(x^2+\frac{1}{x^2}\right)^2+\ldots+\left(x^{12}+\frac{1}{x^{12}}\right)^2\right. \\
& =\left(\omega+\frac{1}{\omega}\right)^2+\left(\omega^2+\frac{1}{\omega^2}\right)^2+\ldots+\left(\omega^{12}+\frac{1}{\omega^{12}}\right)^2 \\
& =\left(\omega+\frac{1}{\omega}\right)^2+\left(\omega^2+\frac{1}{\omega^2}\right)^2+\ldots+\left(\omega^{11}+\frac{1}{\omega^{11}}\right)^2 \\
& =\left(\omega+\omega^2\right)^2+\left(\omega^2+\omega\right)^2+(1+1)^2+\left(\omega+\omega^2\right)^2 \\
& \quad \quad \quad+\left(\omega^2+\omega\right)^2+(1+1)^2+\left(\omega+\omega^2\right)^2 \\
& =8\left(\omega+\omega^2\right)^2+4(1+1)^2 \\
& =8(-1)^2+4\left(2^2=8+16=24\right.
\end{aligned}\)