Mathematics · 2026
JEE Main · 21 January 2026, Shift 1 · Q4
If x^2+x+1=0, then the value of (x+1/x)^4+(x^2+1/x^2)^4+(x^3+1/x^3)^4+…+(x^25+1/(x^25))^4 is:
If $\displaystyle x^2+x+1=0$, then the value of $\displaystyle \left(x+\frac{1}{x}\right)^4+\left(x^2+\frac{1}{x^2}\right)^4+\left(x^3+\frac{1}{x^3}\right)^4+\ldots+\left(x^{25}+\frac{1}{x^{25}}\right)^4$ is :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 145$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.