Question

Factorise. 9th grade maths​

Answer 1

here is ur answer

hope it helps u

Answer 2

$\huge{\bold{\mathfrak{\pink{Solution}}}}$

$\boxed{FACTORISE -:}$

(i)

${8a}^{3} + {b}^{3} + 12 {a}^{2} b + 6a {b}^{2}$

$= (2a)^{3} + (b)^{3} + 3 \times 2a \times b(2a + b)$

$since \: the \: above \: expression \: is \: in \: the \: form \: of \: eqution \: (a + b)^{3}$

$therefore \: the \: answer \: = (2a + b)^{3}$

(ii)

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

$= (4a)^{3} - (3b)^{3} - 3 \times 4a \times 3b(4a - 3b)$

$since \: this \: expression \: is \: in \: the \: form \: of \: (a-b)^{3}$

$therefore \: the \: answer \: = (4a - 3b)^{3}$