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Mathematics · 2026

JEE Main · 22 January 2026, Shift 1 · Q21

Let α=(-1+i √ 3)/2 and β=(-1-i √ 3)/2, i=√ -1. If (7-7 α+9 β)^20+(9+7 α-7 β)^20+(-7+9 α+7 β)^20+(14+7 α+7 β)^20=m^10, then m is ____

Let $\displaystyle \alpha=\frac{-1+i \sqrt{3}}{2}$ and $\displaystyle \beta=\frac{-1-i \sqrt{3}}{2}, i=\sqrt{-1}$. If $\displaystyle (7-7 \alpha+9 \beta)^{20}+(9+7 \alpha-7 \beta)^{20}+(-7+9 \alpha+7 \beta)^{20}+(14+7 \alpha+7 \beta)^{20}=m^{10}$, then $\displaystyle m$ is $\displaystyle \_\_\_\_$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.