Question

8a^3-b^3-4ax+2bx factorise

Answer 1

The answer is given below :

Now,

$8 {a}^{3} - {b}^{3} - 4ax + 2bx \\ \\ = ( {(2a)}^{3} - {b}^{3} ) - 2x(2a - b) \\ \\ = (2a - b)( {(2a)}^{2} + (2a \times b) + {b}^{2} ) - 2x(2a - b) \\ \\ = (2a - b)(4 {a}^{2} + 2ab + {b}^{2} ) - 2x(2a - b) \\ \\ = (2a - b)(4 {a}^{2} + 2ab + {b}^{2} - 2x)$

which is the required factorization.

Thank you for your question.

Answer 2

Hey..Here is your answer u r looking for..
Hey apply Formula of a^3-b^3 in second step......
(2a-b)(4a^2+2ab+b^2)-2x(2a-b)
(2a-b)(4a^2+2ab+b^2-2x)
By mistake I apply formula of a^3+b^3

see the attachment