Factorize :
${(x - y)}^{3} - 8 x^{3}$

(x - y) (7 x^{2} - 4 y + y^{2})

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

- (x + y) (7 x^{2} - 4 x y + y^{2})

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

- (x - y) (7 x^{2} - x y + y^{3})

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

(x + y) (7 x^{2} - x y + y^{2})

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

Open in App

Solution

The correct option is A $- (x + y) (7 x^{2} - 4 x y + y^{2})$
${(x - y)}^{3} - 8 x^{3}$
$= {(x - y)}^{3} - (2 x)^{3}$
Using,
$(a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})$
$= (x - y - 2 x) [(x - y)^{2} + (x - y) (2 x) + (2 x)^{2}]$
$= (- x - y) [(x^{2} + y^{2} - 2 x y + 2 x^{2} - 2 x y + 4 x^{2}]$
$= - (x + y) (7 x^{2} - 4 x y + y^{2})$

Suggest Corrections

Similar questions

The correct factorise form of $(x - y)^{3} - 8 x^{3}$ is _______

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{(x - y)^{2} + 3 x y}] + [\frac{(x + y)^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x 3 - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

Join BYJU'S Learning Program