Jee Main
Complex Numbers
JEE Main 2026 (Online) 21st January Morning Shift
MCQ+4 / -12026
If $x^2+x+1=0$, then the value of $\left(x+\frac{1}{x}\right)^4+\left(x^2+\frac{1}{x^2}\right)^4+\left(x^3+\frac{1}{x^3}\right)^4+\ldots+\left(x^{25}+\frac{1}{x^{25}}\right)^4$ is:
