Factorise: ${(a + b)}^{3} - {(a - b)}^{3}$

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Solution

Given that:
${(a + b)}^{3} - {(a - b)}^{3}$
$[∵ {(a + b)}^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2} a n d {(a - b)}^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}]$
So
$= a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2} - (a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2})$
$= a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2} - a^{3} + b^{3} + 3 a^{2} b - 3 a b^{2}$
$= 2 b^{3} + 6 a^{2} b$
Taking2b common.
$= 2 b (b^{2} + 3 a^{2})$
So
${(a + b)}^{3} - {(a - b)}^{3} = 2 b (b^{2} + 3 a^{2})$

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