Question

. Factorise: 64a³-27b³-144a²b+108ab?
a. (4a – 3b) (4a - 3b) (4a - 3b)
b. (4a + 3b) (4a + 3b) (4a + 3b)
C. (4a + 3b) (4a - 3b)
d. (4a - 3b) (4a - 3b)
​

Answer 1

${ \huge{ \fbox{ \tt{ \orange{Question:-}}}}}$

Factorise the given question.

$64 {a}^{3} - 27 {b}^{3} - 144 {a}^{2}b + 108ab$

${ \huge{ \fbox{ \tt{ \purple{Answer:-}}}}}$

$64 {a}^{3} can \: be \: written \: as \: (4a) {}^{3}$

$27 {b}^{3} can \: be \: written \: as \: (3b) {}^{3}$

So , the factorization is :

$(4a + 3b) {}^{3}$

${ \blue{ \tt{ \underline{ \underline{Verification:-}}}}}$

We know the identity :

$(x + y) {}^{3} = x {}^{3} + y {}^{3} + 3xy(x + y)$

So when we put up the value we get :

$64 {a}^{3} - 27 {b}^{3} - 144 {a}^{2}b + 108ab$

so the answer is option b .

(4a+3b)(4a+3b)(4a+3b) or (4a+3b)^3