If $1, \omega, \omega^2$ are the cube roots of unity then the value of $(x+y)^2+\left(x \omega+y \omega^2\right)^2+\left(x \omega^2+y \omega\right)^2$ is

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Aicharya · Accepted Answer

Given that $1, \mathrm{w}, \mathrm{w}^2$ are the cube roots of unity. $	herefore 1+\mathrm{w}+\mathrm{w}^2=0$ and $\mathrm{w}^3=1$. Now, $(x+y)^2\left(x w+y w^2ight)+\left(x w^2+y wight)^2$ $\begin{aligned} & =x^2+y^2+2 x y+x^2 w^2+y^2 w^4+2 x y w^3 \ & +x^2 w^4+y^2 w^2+2 x y w^3 \ & =x^2+y^2+2 x y+x^2 w^2+y^2 w+2 x y+x^2 w+y^2 w^2+2 x y \ & =x^2+y^2+6 x y+x^2\left(w^2+wight)+y^2\left(w+w^2ight) \ & =x^2+y^2+6 x y-x^2-y^2=6 x y\end{aligned}$

If $1, \omega, \omega^2$ are the cube roots of unity then the value of $(x+y)^2+\left(x \omega+y…

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