If $x = α + β, y = α ω + β ω^{2}, z = α ω^{2} + β ω$ , $ω$ is an imaginary cube root of unity. Then, the value of $x y z$ is.

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Solution

Find the value of $x y z$ :
Given,
$x = α + β$
$y = α ω + β ω^{2}$
$z = α ω^{2} + β ω$
Then,
$\begin{array}{rcl} x y z & = & (α + β) (α ω + β ω^{2}) (α ω^{2} + β ω) \\ = & (α^{2} ω + α β ω^{2} + α β ω + β^{2} ω^{2}) (α ω^{2} + β ω) \\ = & α^{3} + α^{2} β ω + α^{2} β + α β^{2} ω + α^{2} β ω^{2} + α β^{2} + α β^{2} ω^{2} + β^{3} \\ = & α^{3} + β^{3} + α^{2} β (1 + ω + ω^{2}) + α β^{2} (1 + ω + ω^{2}) \\ = & α^{3} + β^{3} + 0 + 0 [∵ if ω is cube root of unity then, 1 + ω + ω^{2} = 0] \\ = & α^{3} + β^{3} \end{array}$
Hence, the correct answer is $\begin{array}{l} α^{3} + β^{3} \end{array}$ .

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