Factorise
$64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$

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Solution

$64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2} = {(4 a)}^{3} - {(3 b)}^{3} - 3 {(4 a)}^{2} (3 b) + 3 (4 a) {(3 b)}^{2} = {(4 a - 3 b)}^{3}$

Hence, factorisation of $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$ is ${(4 a - 3 b)}^{3}$ .

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Similar questions

Factorise each of the following: 64 $a^{3}$ – 27 $b^{3}$ – 144 $a^{2}$ b + 108a $b^{2}$

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}

64 a^{3}

- 27 b^{3}

- 144 a^{2} b

+ 108 a b^{2}

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}

= (4 a - 3 b) (4 a - 3 b) (4 a - 3 b)

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