If $1, \omega, \omega^2$ are the cube roots of unity and $(x+y)(x \omega+$ $\left.y \omega^2\right)\left(x \omega^2+y \omega\right)=f(x, y)$, then $f(2,3)=$

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Since, } f(x, y)=(x+y)\left(x \omega+y \omega^2ight)\left(x \omega^2+y \omegaight) \ & \Rightarrow f(2,3)=(2+3)\left(2 \omega+3 \omega^2ight)\left(2 \omega^2+3 \omegaight) \ & =5\left(4 \omega^3+6 \omega^2+6 \omega^4+9 \omega^3ight) \ & =5\left(4+6 \omega^2+6 \omega+9ight) \ & =5\left(13+6\left(\omega^2+\omegaight)ight)=5(13+6 	imes(-1)) \quad\left[\because \omega^3=1ight] \ & =5 	imes 7=35\end{aligned}$

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

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