Factorise: $a^{3} - 27 b^{3} + 2 a^{2} b - 6 a b^{2}$

(a - b) (a^{2} - 5 a b + 9 b^{2})

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

(a - 3 b) (a^{2} + 5 a b + 9 b^{2})

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

(a - 2 b) (a^{2} + a b + 9 b^{2})

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

(b - a) (a^{2} + 5 b + b^{2})

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

Open in App

Solution

The correct option is B $(a - 3 b) (a^{2} + 5 a b + 9 b^{2})$
$a^{3} - 27 b^{3} + 2 a^{2} b - 6 a b^{2}$
$= (a - 3 b) (a^{2} + a (3 b) + (3 b)^{2} + 2 a b (a - 3 b)$
$= (a - 3 b) (a^{2} + 3 a b + 9 b^{2} + 2 a b)$
$= (a - 3 b) (a^{2} + 5 a b + 9 b^{2})$

Suggest Corrections

Similar questions

a^{3} - 27 b^{3} + 2 a^{2} b - 6 a b^{2}

Factorise Completely : $4 a^{2} - 12 a b + 9 b^{2} + 4 a - 6 b$

25 a^{2} - 9 b^{2}

3 - 12 (a - b)^{2}

4 a^{2} - 9 b^{2} - 2 a - 3 b

If $(8 a^{3} + 27 b^{3}) = 0$

then

$(4 a^{2} + 9 b^{2}) =$

Join BYJU'S Learning Program