The product of $(4 x - 3 y)$ and $(16 x^{2} + 12 x y + 9 y^{2})$ is

(4 x + 9 y)^{3}

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

(16 x^{2} - 6 x y + 9 y^{2})^{2}

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

64 x^{3} - 27 y^{3}

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

None of these

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

Open in App

Solution

The correct option is C $64 x^{3} - 27 y^{3}$
$(4 x - 3 y) (16 x^{2} + 12 x y + 9 y^{2})$
$= (4 x - 3 y) [(4 x)^{2} + (4 x) (3 y) + (3 y)^{2}]$
$= (4 x)^{3} - (3 y)^{3}$ ... $[(a - b) (a^{2} + a b + b^{2}) = a^{3} - b^{3}]$
$= 64 x^{3} - 27 y^{3}$

Suggest Corrections

Similar questions

Q. Factorise 4x^2+9y^2-12xy

Q. The product of

4 x y

and

4 x + 3 y

16 x^{2} y + 12 x y^{2}

Find the following products:

(i) (3x + 2y + 2z) (9x² + 4y² + 4z² − 6xy − 4yz − 6zx)

(ii) (4x − 3y + 2z) (16x² + 9y² + 4z² + 12xy + 6yz − 8zx)

(iii) (2ab − 3b − 2c) (4a² + 9b² +4c² + 6 ab − 6 bc + 4ca)

(iv) (3x − 4y + 5z) (9x² +16y²+ 25z² + 12xy −15zx + 20yz)

Q. Find the centre, eccentricity, foci and directrices of the hyperbola
(i) 16x² − 9y² + 32x + 36y − 164 = 0
(ii) x² − y² + 4x = 0
(iii) x² − 3y² − 2x = 8.

$Find the centre,eccentricity,foci and directrices of the hyperbola (i) 16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0 (i i) x^{2} - y^{2} + 4 x = 0 (i i i) x^{2} - 3 y^{2} - 2 x = 8$

Join BYJU'S Learning Program

The product of (4x−3y) and (16x2+12xy+9y2) is

The correct option is C 64x3−27y3(4x−3y)(16x2+12xy+9y2)=(4x−3y)[(4x)2+(4x)(3y)+(3y)2]=(4x)3−(3y)3 ...[(a−b)(a2+ab+b2)=a3−b3]=64x3−27y3

The product of $(4 x - 3 y)$ and $(16 x^{2} + 12 x y + 9 y^{2})$ is

The correct option is C $64 x^{3} - 27 y^{3}$
$(4 x - 3 y) (16 x^{2} + 12 x y + 9 y^{2})$
$= (4 x - 3 y) [(4 x)^{2} + (4 x) (3 y) + (3 y)^{2}]$
$= (4 x)^{3} - (3 y)^{3}$ ... $[(a - b) (a^{2} + a b + b^{2}) = a^{3} - b^{3}]$
$= 64 x^{3} - 27 y^{3}$