If $1, ω, ω^{2}$ are cube the roots of unity, then find the values of the following.
i) ${(a + b)}^{3} + {(a ω + b ω^{2})}^{3} + {(a ω^{2} + b ω)}^{3}$
ii) ${(a + 2 b)}^{2} + {(a ω^{2} + 2 b ω)}^{2} + {(a ω + 2 b ω^{2})}^{2}$

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Solution

Since, w.k.t $w^{3} = 1, 1 + w + w^{2} = 0$

(i)

$(a + b)^{3} + (a w + b w^{2})^{3} + (a w^{2} + b w)^{3}$

$= a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} + a^{3} + 3 a^{2} b w + 3 a b^{2} w^{2} + b^{3} + a^{3} + 3 a^{2} b w^{2} + 3 a b^{2} w + b^{3}$

$= 3 a^{3} + 3 b^{3} + 3 a^{2} b (1 + w + w^{2}) + 3 a b^{2} (1 + w + w^{2})$

$= 3 a^{3} + 3 b^{3}$

$= 3 (a^{3} + b^{3})$

(ii)

$(a + 2 b)^{2} + (a w^{2} + 2 b w)^{2} + (a w + 2 b w^{2})^{2}$

$= a^{2} + 4 a b + 4 b^{2} + a^{2} w + 4 a b + 4 b^{2} w^{2} + a^{2} w^{2} + 4 a b + 4 b^{2} w$

$= a^{2} (1 + w + w^{2}) + 4 b^{2} (1 + w + w^{2}) + 8 a b$

$= 8 a b$

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