Factorise the following: $a^{6} - b^{6}$

(a + b) (a - b) (a^{2} + b^{2} + a b) (a^{2} - b^{2} + a b)

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

(a + b) (a - b) (a^{2} + b^{2} + a b) (a^{2} + b^{2} - a b)

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

(a + b) (a - b) (a^{2} - b^{2} - a b) (a^{2} + b^{2} - a b)

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

None of these

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

Open in App

Solution

The correct option is D $(a + b) (a - b) (a^{2} + b^{2} + a b) (a^{2} + b^{2} - a b)$
$a^{6} - b^{6}$
Using, $a^{2} - b^{2} = (a + b) (a - b)$
$= (a^{3})^{2} - (b^{3})^{2}$
$= (a^{3} + b^{3}) (a^{3} - b^{3})$
Using, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}), a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$
$= (a + b) (a^{2} - a b + b^{2}) (a - b) (a^{2} + a b + b^{2})$
$= (a + b) (a - b) (a^{2} - a b + b^{2}) (a^{2} + a b + b^{2})$

Suggest Corrections

Similar questions

a^{5} b - a b^{5}

a^{6} - b^{6}

On Resolving $a^{6} - b^{6}$ into factors give $(a + b) (a - b) (a^{2} - a b + b^{2}) (a^{2} + a b + b^{2})$ .

$The product (a + b) (a - b) (a^{2} - a b + b^{2}) (a^{2} + a b + b^{2})$ is equal to

Join BYJU'S Learning Program