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  4. If x=a+b+c, y=a α+b β+c and z=a β+b α+c, where α and β are complex cube roots of unity, then x y z=

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Q. If $x=a+b+c, y=a \alpha+b \beta+c$ and $z=a \beta+b \alpha+c$, where $\alpha$ and $\beta$ are complex cube roots of unity, then $x y z=$

Complex Numbers and Quadratic Equations

Solution:

$x=a+b+c, y=a \omega+b \omega^2+c, z=a \omega^2+b \omega+c$
$y z=\left(a \omega+b \omega^2+c\right)\left(a \omega^2+b \omega+c\right)$
$=a^2+b^2+c^2+a b\left(\omega^4+\omega^2\right)+b c\left(\omega^2+\omega\right)+c a\left(\omega^2+\omega\right)$
$=a^2+b^2+c^2+a b\left(\omega^2+\omega\right)+b c\left(\omega^2+\omega\right)+c a\left(\omega^2+\omega\right)$
$=a^2+b^2+c^2-a b-b c-c a \left\{\omega^2+\omega+1=0\right\}$
$x y z=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=a^3+b^3+c^3-3 a b c$