Let {omega _n} = cos left( {frac{{2pi }}{n}} | vedclass

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(C) Given ${\omega _n} = \cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}} \right)$.For $n=3$,${\omega _3} = \cos \left( {\fr

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${x^2} + {y^2} + {z^2} - yz - zx - xy$

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(C) Given ${\omega _n} = \cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}} \right)$.For $n=3$,${\omega _3} = \cos \left( {\frac{{2\pi }}{3}} \right) + i\sin \left( {\frac{{2\pi }}{3}} \right) = -\frac{1}{2} + i\frac{{\sqrt{3}}}{2} = \omega$,where $\omega$ is the complex cube root of unity.Similarly,${\omega _3}^2 = \cos \left( {\frac{{4\pi }}{3}} \right) + i\sin \left( {\frac{{4\pi }}{3}} \right) = -\frac{1}{2} - i\frac{{\sqrt{3}}}{2} = {\omega ^2}$.Now,the expression is $(x + y\omega + z{\omega ^2})(x + y{\omega ^2} + z\omega)$.Expanding this product:$= x(x + y{\omega ^2} + z\omega) + y\omega(x + y{\omega ^2} + z\omega) + z{\omega ^2}(x + y{\omega ^2} + z\omega)$$= {x^2} + xy{\omega ^2} + xz\omega + xy\omega + {y^2}{\omega ^3} + yz{\omega ^2} + xz{\omega ^2} + yz{\omega ^4} + {z^2}{\omega ^3}$Since ${\omega ^3} = 1$ and ${\omega ^4} = \omega$:$= {x^2} + {y^2} + {z^2} + xy({\omega ^2} + \omega) + yz({\omega ^2} + \omega) + xz({\omega ^2} + \omega)$Since $1 + \omega + {\omega ^2} = 0$,we have $\omega + {\omega ^2} = -1$.$= {x^2} + {y^2} + {z^2} + xy(-1) + yz(-1) + xz(-1)$$= {x^2} + {y^2} + {z^2} - xy - yz - zx$.

Let ${\omega _n} = \cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}} \right)$ and ${i^2} = -1$. Then $(x + y{\omega _3} + z{\omega _3}^2)(x + y{\omega _3}^2 + z{\omega _3})$ is equal to:

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