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If $\omega$ is cube root of $1$ and $\alpha,\beta,\gamma$ are cube roots of $p$ , then for any real numbers $x,y,z$the value of $\large\frac{x\alpha+y\beta+z\gamma}{x\beta+y\gamma+z\alpha}$ is ?

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Given: $\alpha,\beta,\gamma$ are roots of $p$ and $\omega$ is cube root of $1$
$\Rightarrow\:\alpha=p^{\large\frac{1}{3}}$$,\:\:\beta=\omega p^{\large\frac{1}{3}}$$,\:\:\gamma=\omega^2p^{\large\frac{1}{3}}$
Let $z=\large\frac{x\alpha+y\beta+z\gamma}{x\beta+y\gamma+z\alpha}$
$=\large\frac{xp^{\large\frac{1}{3}}+y\omega p^{\large\frac{1}{3}}+z\omega^2 p^{\large\frac{1}{3}}}{x\omega p^{\large\frac{1}{3}}+y\omega^2 p^{\large\frac{1}{3}}+zp^{\large\frac{1}{3}}}$
$=\large\frac{x+y\omega+z\omega^2}{x\omega+y\omega^2+z}$
$=\large\frac{\omega^2(x\omega+y\omega^2+z)}{x\omega+y\omega^2+z}$$=\omega^2$
Similarly by taking $\beta=\omega^2p^{\large\frac{1}{3}}$ and $\gamma=\omega p^{\large\frac{1}{3}}$
we get $z=\omega$
answered Jul 28, 2013 by rvidyagovindarajan_1
 

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