Factorization of
$(a - b)^{3} + (b - c)^{3} + (c - a)^{3}$ is:

(a - b) (a - c)

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

(a - b) (b - c) (c - a)

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

3 (a - b) (b - c) (c - a)

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

3^{2} (a - b) (b - c) (c - a)

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

Open in App

Solution

The correct option is C $3 (a - b) (b - c) (c - a)$
Let $x = a - b; y = b - c; z = c - a$ .
On adding, $x + y + z = 0$
$∴ x^{3} + y^{3} + z^{3} = 3 x y z$
Substituting
$x = (a - b); y = (b - c)$ and $z = (c - a)$
$(a - b)^{3} + (b - c)^{3} + (c - a)^{3}$
$= 3 (a - b) (b - c) (c - a)$

Suggest Corrections

Similar questions

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

\frac{(a + b)^{3} + (b + c)^{3} + (c + a)^{3} - 3 (a + b) (b + c) (c + a)}{3 (a^{3} + b^{3} + c^{3} - 3 a b c)}

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)

Join BYJU'S Learning Program