If $1, \omega, \omega^2$ are the cube roots of unity, $\mathrm{k}$ is positive integer and $\left(1-\omega+\omega^2\right)^{3 \mathrm{k}}+\left(1-\omega^2+\omega\right)^{3 \mathrm{k}}=\left(1-\omega+\omega^2\right)^{3 \mathrm{k}+1}+$ $\left(1+\omega-\omega^2\right)^{3 \mathrm{k}+1}$, then…

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If $1, \omega, \omega^2$ are the cube roots of unity, $\mathrm{k}$ is positive integer and $\left(1-\omega+\omega^2\right)^{3 \mathrm{k}}+\left(1-\omega^2+\omega\right)^{3 \mathrm{k}}=\left(1-\omega+\omega^2\right)^{3 \mathrm{k}+1}+$ $\left(1+\omega-\omega^2\right)^{3 \mathrm{k}+1}$, then $\mathrm{k}=$

Aicharya · Accepted Answer

$\begin{aligned} & 	ext {}\left(1-\omega+\omega^2ight)^{3 k}+\left(1-\omega^2+\omegaight)^{3 k} \ & =\left(1-\omega+\omega^2ight)^{3 k+1}+\left(1+\omega-\omega^2ight)^{3 k+1} \ & \Rightarrow(-\omega-\omega)^{3 k}+\left(-\omega^2-\omega^2ight)^{3 k} \ & =(-\omega-\omega)^{3 k+1}+\left(-\omega^2-\omega^2ight)^{3 k+1} \ & \Rightarrow(-2)^{3 k}(\omega)^3+(-2)^{3 k}\left(\omega^2ight)^3 \ & =(-2 \omega) \cdot(-2)^{3 k}(\omega)^{3 k}+\left(-2 \omega^2ight) \cdot(-2)^{3 k} \omega^{3 k} \ & \Rightarrow 1+1=-2 \omega-2 \omega^2 \ & \Rightarrow 2=-2\left(\omega+\omega^2ight) \ & \Rightarrow 1=-1(-1) \Rightarrow 1=1 \end{aligned}$ $	herefore \quad$ The given statement is true for all $k \in N$ $	herefore \quad k=r$, where $r \in N$

If $1, \omega, \omega^2$ are the cube roots of unity, $\mathrm{k}$ is positive integer and…

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