Question

Factorise a³ - b³ + 4a-4b

Answer 1

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

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)`