Factorise:

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

(a + b) (7 a^{2} + 7 b^{2} + 13 a b)

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(a - b) (7 a^{2} + 7 b^{2} + 13 a b)

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(a - b) (7 a^{2} + 7 b^{2} + 11 a b)

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(a - b) (7 a^{2} + 7 b^{2} + 12 a b)

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Solution

The correct option is A $(a - b) (7 a^{2} + 7 b^{2} + 13 a b)$
${(2 a + b)}^{3} - {(a + 2 b)}^{3}$
Using $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ , we get
$= (2 a + b - a - 2 b) ((2 a + b)^{2} + (2 a + b) (a + 2 b) + (a + 2 b)^{2})$
$= (a - b) (4 a^{2} + 2 (2 a) b + b^{2} + 2 a^{2} + 4 a b + a b + 2 b^{2} + a^{2} + 4 b^{2} + 4 a b)$
$= (a - b) (7 a^{2} + 7 b^{2} + 13 a b)$

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Similar questions

Factorise : $7 b^{2} - 8 b + 1$

a = \frac{1}{2 - \sqrt{3}}, b = \frac{1}{2 + \sqrt{3}}

7 a^{2} + 11 a b - 7 b^{2}

\frac{3 a + 7 b}{3 a - 7 b} = \frac{4}{3}

\frac{3 a^{2} - 7 b^{2}}{3 a^{2} + 7 b^{2}}

\frac{3 a + 7 b}{3 a - 7 b} = \frac{4}{3}

\frac{3 a^{2} - 7 b^{2}}{3 a^{2} + 7 b^{2}}

{(4 a - 7 b)}^{2} - {(3 a)}^{2}

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