Question

Factorise a³ - b³ + 4a-4b

Answer 1

Answer:

$a^3-b^3+4a-4b:\:\left(a-b\right)\left(a^2+ab+b^2+4\right)$

Step-by-step explanation:

Given the expression

$a^3-b^3+4a-4b$

solving the expression

$a^3-b^3+4a-4b$

as

• $a^3-b^3$

$\mathrm{Apply\:Difference\:of\:Cubes\:Formula:\:}x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)$

$a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$

also

• $4a-4b$

$\mathrm{Factor\:out\:common\:term\:}4$

$=4\left(a-b\right)$

so the expression becomes

$=\left(a-b\right)\left(a^2+ab+b^2\right)+4\left(a-b\right)$

$\mathrm{Factor\:out\:common\:term\:}\left(a-b\right)$

$=\left(a-b\right)\left(a^2+ab+b^2+4\right)$

Therefore,

$a^3-b^3+4a-4b:\:\left(a-b\right)\left(a^2+ab+b^2+4\right)$