Factorise : $a^{3} - 8 b^{3} - 64 c^{3} - 24 a b c$

(a + 2 b - 4 c) (a^{2} + 4 b^{2} + 16 c^{2} + 2 a b + 8 b c + 4 a c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(a - 2 b + 4 c) (a^{2} + 4 b^{2} + 16 c^{2} + 2 a b + 8 b c + 4 a c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(a + 2 b - 4 c) (a^{2} + 4 b^{2} + 16 c^{2} + 2 a b - 8 b c + 4 a c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(a - 2 b - 4 c) (a^{2} + 4 b^{2} + 16 c^{2} + 2 a b - 8 b c + 4 a c)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $(a - 2 b - 4 c) (a^{2} + 4 b^{2} + 16 c^{2} + 2 a b - 8 b c + 4 a c)$
We know that,
$a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 a b c$
Thus,
$a^{3} - 8 b^{3} - 64 c^{3} - 24 a b c$
Here, $a = a$ , $b = - 2 b$ and $c = - 4 c$
$= (a - 2 b - 4 c) [a^{2} + (- 2 b)^{2} + (- 4 c)^{2} - a (- 2 b) - (- 2 b) (- 4 c) - (- 4 c) a]$
$= (a - 2 b - 4 c) (a^{2} + 4 b^{2} + 16 c^{2} + 2 a b - 8 b c + 4 c a)$

Suggest Corrections

Similar questions

9 a^{2} + 4 b^{2} + 16 c^{2} + 12 a b - 16 b c - 24 c a

Factorise $(a^{3} - 8 b^{3})$ .

(2 b - a) (2 b + c)

4 b^{2} - 2 a b + 2 b c - a c

Expand the following, using suitable identities: ${(\frac{1}{4} a - \frac{1}{2} b + 1)}^{2}$

$a^{2} + 2 a b + 4 b^{2}$ a perfect square trinomial

Join BYJU'S Learning Program