# (x^3-1)/(x^2+1)div (9x^2+9)/(x^2-x)=

$\frac{{x}^{3}-1}{{x}^{2}+1}÷\frac{9{x}^{2}+9}{{x}^{2}-x}=$
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To divide rational expressions, we use the following property offractions:
$\frac{A}{B}÷\frac{C}{D}=\frac{A}{B}\ast \frac{D}{C}$ (Note: this is also called inverting andmultiplying).
So, $\frac{{x}^{3}-1}{{x}^{2}+1}÷\frac{9{x}^{2}+9}{{x}^{2}-x}=\frac{{x}^{3}-1}{{x}^{2}+1}\ast \frac{{x}^{2}-x}{9{x}^{2}+9}$
$=\frac{\left(x-1\right)\left({x}^{2}+x+1\right)}{{x}^{2}+1}\ast \frac{x\left(x-1\right)}{9\left({x}^{2}+1\right)}←$ Factor what you can here.
$=\frac{x\left(x-1{\right)}^{2}\left({x}^{2}+x+1\right)}{9\left({x}^{2}+1{\right)}^{2}}$