If W is the complex cube root of unity then prove that $(2 + 5 W + 2 W^{2})^{6} = (2 + 2 W + 5 W^{2})^{6} = 729$ .

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Solution

We know that,
$1 + W + W^{2} = 0 W^{3} = 1$
(Properties of Cube Root of unity)

Now,
$(2 + 5 W + 2 W^{2})^{6}$
$= (2 + 2 W + 2 W^{2} + 3 W)^{6}$
$= (2 (1 + W + W^{2}) + 3 W)^{6}$
$= (3 W + 0)^{6} = (3 W)^{6} = 3^{6} (W^{3})^{2}$
$= 3^{6} * 1^{2}$
$= 729$

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