Find the product of $(2 x - 3 y)$ and $(4 x^{2} + 9 y^{2} + 6 x y)$ .

$8 x^{3} - 27 y^{3}$

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

$2 x^{3} - 3 y^{3}$

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

$(2 x - 3 y)^{3}$

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

$(8 x - 27 y)^{3}$

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

Open in App

Solution

The correct option is A
$8 x^{3} - 27 y^{3}$

We have to find the product: $(2 x - 3 y) \times (4 x^{2} + 9 y^{2} + 6 x y)$ .

Observe the expression carefully, it is the expansion of difference of cube of two terms.

Using the identity: $a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b),$ we get

$(2 x - 3 y) (4 x^{2} + 9 y^{2} + 6 x y) = (2 x)^{3} - (3 y)^{3} = 8 x^{3} - 27 y^{3}$

Suggest Corrections

Similar questions

(2 x - 3 y + 5 z)

(4 x^{2} + 9 y^{2} + 25 z^{2} + 6 x y + 15 y z - 10 x z)

(4 x^{2} + 9 y^{2}

4 x^{2} + 9 y^{2} = 40, x y = 2

2 x + 3 y

(2 x - 3 y) = 10

x y = 8

4 x^{2} + 9 y^{2}

Join BYJU'S Learning Program