Simplify: $\dfrac{x+3}{x^{3}-1}\div\dfrac{3x+9}{x^{2}+x+1}$
$\dfrac{1}{3x+1}$
3x+1
3x-3
$\dfrac{1}{3x-3}$
Explanation:
By applying formula:
$x^3-y^3 = (x-y)(x^2+xy+y^2)$
By applying formula:
$x^3-y^3 = (x-y)(x^2+xy+y^2)$
$\dfrac{x+3}{(x-1)(x^{2}+x+1)}\times\dfrac{x^{2}+x+1}{3(x+3)}$
$=\dfrac{1}{3x-3}$