If $1, w, w^{2}$ are the cube roots of unity, prove that ${(1 - w + w^{2})}^{6} + {(1 - w^{2} + w)}^{6} = 128$ .

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Solution

${(1 - w + w^{2})}^{6} + {(1 - w^{2} + w)}^{6}$ .
$= (- ω - ω)^{6} + (- ω^{2} - ω^{2})^{6}$ , since $ω + ω^{2} = - 1$
$= 2^{6} ω^{6} + 2^{6} ω^{12}$
$= 2^{6} + 2^{6} = 64 + 64 = 128$ , since $ω^{multiple of 3} = 1$

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