Let \[{{\omega }_{n}}=\cos \left( \frac{2\pi }{n} \right)+i\,\sin \left( \frac{2\pi }{n} \right)\,,\,{{i}^{2}}=-1\], then \[(x+y{{\omega }_{3}}+z{{\omega }_{3}}^{2})\] \[(x+y{{\omega }_{3}}^{2}+z{{\omega }_{3}})\] is equal to [AMU 2001]

\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+yz+zx+xy\]

\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}\]

\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-yz-zx-xy\]\[\]

Let \[{{\omega }_{n}}=\cos \left( \frac{2\pi }{n} \right)+i\,\sin \lef

1.	Let \[{{\omega }_{n}}=\cos \left( \frac{2\pi }{n} \right)+i\,\sin \left( \frac{2\pi }{n} \right)\,,\,{{i}^{2}}=-1\], then \[(x+y{{\omega }_{3}}+z{{\omega }_{3}}^{2})\] \[(x+y{{\omega }_{3}}^{2}+z{{\omega }_{3}})\] is equal to [AMU 2001]
A.	0
B.	\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}\]
C.	\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-yz-zx-xy\]\[\]
D.	\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+yz+zx+xy\]
Answer» D. \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+yz+zx+xy\]

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Let \[{{\omega }_{n}}=\cos \left( \frac{2\pi }{n} \right)+i\,\sin \left( \frac{2\pi }{n} \right)\,,\,{{i}^{2}}=-1\], then \[(x+y{{\omega }_{3}}+z{{\omega }_{3}}^{2})\] \[(x+y{{\omega }_{3}}^{2}+z{{\omega }_{3}})\] is equal to [AMU 2001]

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