Answer:
[tex]a^3-b^3+4a-4b:\:\left(a-b\right)\left(a^2+ab+b^2+4\right)[/tex]
Step-by-step explanation:
Given the expression
[tex]a^3-b^3+4a-4b[/tex]
solving the expression
[tex]a^3-b^3+4a-4b[/tex]
as
[tex]\mathrm{Apply\:Difference\:of\:Cubes\:Formula:\:}x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)[/tex]
[tex]a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)[/tex]
also
[tex]\mathrm{Factor\:out\:common\:term\:}4[/tex]
[tex]=4\left(a-b\right)[/tex]
so the expression becomes
[tex]=\left(a-b\right)\left(a^2+ab+b^2\right)+4\left(a-b\right)[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}\left(a-b\right)[/tex]
[tex]=\left(a-b\right)\left(a^2+ab+b^2+4\right)[/tex]
Therefore,
[tex]a^3-b^3+4a-4b:\:\left(a-b\right)\left(a^2+ab+b^2+4\right)[/tex]
