Question

a²-4b²+a³-8b³-(a-2b)²;
Factorise the given expression completely. ​

Answer 1

Step-by-step explanation:

click on the photo and hope you like it

Answer 2

The factorise the given expression $a^2-4b^2+a^3-8b^3-(a-2b)^2$ is $(a-2b)(4b+a^{2}+2ab+4b^{2}).$

Step-by-step explanation:

We have,

$a^2-4b^2+a^3-8b^3-(a-2b)^2$

To factorise the given expression=?

$=(a^2-4b^2)+(a^3-8b^3)-(a-2b)^2$

$=[a^2-(2b)^2]+[a^3-(2b)^3]-(a-2b)^2$

$=(a+2b)(a-2b)+(a-2b)(a^{2}+2ab+4b^{2})-(a-2b)^2$

Using identities,

$a^{2}-b^{2} =(a+b)(a-b)$ and

$a^{3}-b^{3} =(a-b)(a^{2}+ab+b^{2})$

Taking (a-2b) as common, we get

$=(a-2b)[(a+2b)+(a^{2}+2ab+4b^{2})-(a-2b)]$

$=(a-2b)(a+2b+a^{2}+2ab+4b^{2}-a+2b)$

$=(a-2b)(4b+a^{2}+2ab+4b^{2})$

$=(a-2b)(4b+a^{2}+2ab+4b^{2})$

Hence, the factorise the given expression $a^2-4b^2+a^3-8b^3-(a-2b)^2$ is $(a-2b)(4b+a^{2}+2ab+4b^{2}).$