Factorise: $(7 a^{3} + 56 b^{3})$ .

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Solution

$7 a^{3} + 56 b^{3} = 7 (a^{3} + 8 b^{3}) = 7 [{(a)}^{3} + {(2 b)}^{3}] = 7 (a + 2 b) [{(a)}^{2} + {(2 b)}^{2} - a \times 2 b] = 7 (a + 2 b) (a^{2} + 4 b^{2} - 2 a b)$

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