Question

how to factorise 36a^3b - 4ab^3

Answer 1

Answer:

$\boxed{\sf 4ab(3a + b)(3a - b)}$

Step-by-step explanation:

$\sf Factor \: the \: following: \\ \sf \implies 36 {a}^{3} b - 4a {b}^{3} \\ \\ \sf Factor \: 4 ab \: out \: of \: 36 {a}^{3} b - 4a {b}^{3} : \\ \sf \implies 4ab(9 {a}^{2} - {b}^{2} ) \\ \\ \sf 9 {a}^{2} - {b}^{2} = {(3a)}^{2} - {b}^{2} : \\ \sf \implies 4ab({(3a)}^{2} - {b}^{2}) \\ \\ \sf Factor \: the \: difference \: of \: two \: squares. \\ \sf {(3a)}^{2} - {b}^{2} = (3a + b)(3a - b) : \\ \sf \implies 4ab(3a + b)(3a - b)$

Answer 2

Answer:

Step-by-step explanation:

QUESTION : Factorize 36a^3b - 4ab^3

Step1 : Split everything.

            so, 2*2*3*3*a*a*a*b - 2*2*a*b*b*b

HINT: ( you can do this step by taking the HCF of 36 and 4 . Then split everything.)

Step2 : Take out the common ones.

            2*2*3*3*a*a*a*b      -       2*2*a*b*b*b

           = 2*2*a*b     -     2*2*a*b

           = 4ab          -         4ab

 Therefore the common is 4ab.

Hint: ( Common numbers are underlines for extra help)

Step3 : Keep the common outside the bracket {eg. 4ab(......) } and then keep the other numbers inside the bracket. (Other numbers means the numbers which were not common or which we didn't underline and then you multiply the non-common numbers and put them in bracket)

  =  4ab( 9a^2 - b^2 )

  SINCE the formula is a^2 - b^2 = (a+b) (a-b) , we will take HCF of 9 which will give you 3. After minus sign there is no other number ,so we don't need to factorize anything there.

Instead of 9 we will put three in both of the brackets. a^ and b^2 will change to a and b. Then your answer will look like :

=4ab(3a+b) (3a-b) will be your final answer.

Was it helpful ? Give a heart <3