Factorise the following:
$64 a^{3} + 21 b^{3}$

Open in App

Solution

$64 a^{3} + 27 b^{3}$
$= (4 a)^{3} + (3 b)^{3}$
$= (4 a + 3 b) [(4 a)^{2} - 4 a .3 b + (3 b)^{2}]$
$[∵ a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})]$
$= (4 a + 3 b) [16 a^{2} - 12 a b + 9 b^{2}]$

Suggest Corrections

Similar questions

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

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

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

Factorise
$64 a^{3} - 343$

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

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

Join BYJU'S Learning Program