Question

27a^3+b^3/ 3ab+b^2

Answer 1

\(\dfrac{27a^3+b^3}{3ab+b^2}\\ =\dfrac{\left(3a+b\right)\left(9a^2-3ab+b^2\right)}{b\left(3a+b\right)}\\ =\dfrac{9a^2-3ab+b^2}{b}\)

Answer 2

\(=\dfrac{\left(3a+b\right)\left(9a^2-3ab+b^2\right)}{b\left(3a+b\right)}=\dfrac{9a^2-3ab+b^2}{b}\)