Factorise:
$a^{3} - b^{3} - c^{3} - 3 a b c$

Open in App

Solution

We know the identity $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

Using the above identity taking $a = a, b = - b$ and $c = - c$ , the equation $a^{3} - b^{3} - c^{3} - 3 a b c$ can be factorised as follows:

$a^{3} - b^{3} - c^{3} - 3 a b c = (a^{3}) + (- b)^{3} + (- c)^{3} - 3 (a) (- b) (c)$
$= [a + (- b) + (- c)] [a^{2} + (- b)^{2} + (- c)^{2} - (a \times - b) - (- b \times - c) - (- c \times a)]$
$= (a - b - c) (a^{2} + b^{2} + c^{2} + a b - b c + c a)$

Hence, $a^{3} - b^{3} - c^{3} - 3 a b c = (a - b - c) (a^{2} + b^{2} + c^{2} + a b - b c + c a)$

Suggest Corrections

Similar questions

a^{3} + 8 b^{3} + c^{3} - 6 a b c

a^{3} + b^{3} + c^{3} - 3 a b c = \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]

a^{3} + b^{3} + c^{3} - 3 a b c

a^{3} + b^{3} + c^{3} = 3 a b c

1^{3} + b^{3} + 8 c^{3} - 6 b c

a^{3} + b^{3} + c^{3} - 3 a b c = \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]

a^{3} + b^{3} + c^{3} - 3 a b c

x^{3} + y^{3} + z^{3} - 3 x y z

A^{3} + B^{3} + C^{3} - 3 A B C

Join BYJU'S Learning Program