Question

a2-4b2+a3-8b3-(a-2b)2
Factorise

Answer 1

Answer:

(a+2b)+(a+2b)(a^2-2ab+4b^2)

=(a+2b)(1+a^2-2ab+4b^2)

a^3+b^3=(a+b)(a^2-ab+b^2). since a^3+8b^3=

a^3+(2b)^3

Answer 2

Thus,  the factorisation of $a^2-4b^2+a^3-8b^3-(a-2b)^2$ is equal to$(a-2b)(a^{2}+4b^2+2ab+4b)$.

Step-by-step explanation:

We have,

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

To find, the factorisation of $a^2-4b^2+a^3-8b^3-(a-2b)^2$ = ?

∴$a^2-4b^2+a^3-8b^3-(a-2b)^2$

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

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

Using the algebraic identity,

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

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

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

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

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

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

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

∴ The factorisation of $a^2-4b^2+a^3-8b^3-(a-2b)^2$ =$(a-2b)(a^{2}+4b^2+2ab+4b)$

Thus,  the factorisation of $a^2-4b^2+a^3-8b^3-(a-2b)^2$ is equal to$(a-2b)(a^{2}+4b^2+2ab+4b)$.