If $\mathrm{w}=\frac{-1+\mathrm{i} \sqrt{3}}{2}$, where $\mathrm{i}=\sqrt{-1}$, then the value of $\left(3+w+3 w^2\right)^4$ is

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Aicharya · Accepted Answer

$\omega$ is a complex cube root of unity $\begin{array}{ll} 	herefore \quad & \omega=1 ...(i)\ 	herefore \quad & 1+\omega+\omega^2=0, \ 	herefore \quad & \left(3+\omega+3 \omega^2ight)^4 ...(ii)\ & =(3+\omega+3(-1-\omega))^4 \ & =(3+\omega-3-3 \omega)^4 \ & =(-2 \omega)^4 \ & =16 \omega^4 \ & =16 \omega^3 	imes \omega \ & =16 \omega \end{array}$ ...[from (ii)] ...[from (i)]

If $\mathrm{w}=\frac{-1+\mathrm{i} \sqrt{3}}{2}$, where $\mathrm{i}=\sqrt{-1}$, then the value of…

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