Factorise each of the following
$1 - 64 a^{3} - 12 a + 48 a^{2}$

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Solution

$1 - 64 a^{3} - 12 a + 48 a^{2}$
= $(1)^{3} - (4 a)^{3} - 3 (1)^{2} (4 a) + 3 (1) (4 a)^{2}$
Suitable identities is $x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2} = (x - y)^{3}$
$∴ (1)^{3} - (4 a)^{3} - 3 (1)^{2} (4 a) + 3 (1) (4 a)^{2}$
= $(1 - 4 a)^{3}$

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1 - 64 a^{3} - 12 a + 48 a^{2}

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Question 32
Factorise the following:
(i) $1 - 64 a^{3} - 12 a + 48 a^{2}$

(ii) $8 p^{3} + \frac{12}{5} p^{2} + \frac{6}{25} p + \frac{1}{125}$

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

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