If 1, omega, omega^2 are the cube roots of un | vedclass

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(C) Given that $1, \omega, \omega^2$ are the cube roots of unity. $	herefore 1+\omega+\omega^2=0$ and $\omega^3=1$. Expanding the given expression: $

Vedclass · Accepted Answer

$6xy$

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(C) Given that $1, \omega, \omega^2$ are the cube roots of unity. $	herefore 1+\omega+\omega^2=0$ and $\omega^3=1$. Expanding the given expression: $(x+y)^2+(x\omega+y\omega^2)^2+(x\omega^2+y\omega)^2$ $= (x^2+y^2+2xy) + (x^2\omega^2+y^2\omega^4+2xy\omega^3) + (x^2\omega^4+y^2\omega^2+2xy\omega^3)$ $= x^2+y^2+2xy + x^2\omega^2+y^2\omega+2xy + x^2\omega+y^2\omega^2+2xy$ $= x^2(1+\omega+\omega^2) + y^2(1+\omega+\omega^2) + 6xy$ Since $1+\omega+\omega^2=0$,we have: $= x^2(0) + y^2(0) + 6xy = 6xy$.

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

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