Question

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

Answer 1

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

Step-by-step explanation:

We have,

To factorise the given expression = ?

∴ $a^2-4b^2+a^3-8b^3-(a-2b)^2$

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

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

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

Using algebraic identity,

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

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

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

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

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

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

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

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

Answer 2

Step-by-step explanation: