Factor out the common term \({x}^{2}\).
\[\frac{\frac{{x}^{2}(x-2)}{3x+3}}{\frac{{x}^{2}-4}{3{x}^{2}+9x+6}}n\]
Factor out the common term \(3\).
\[\frac{\frac{{x}^{2}(x-2)}{3(x+1)}}{\frac{{x}^{2}-4}{3{x}^{2}+9x+6}}n\]
Rewrite \({x}^{2}-4\) in the form \({a}^{2}-{b}^{2}\), where \(a=x\) and \(b=2\).
\[\frac{\frac{{x}^{2}(x-2)}{3(x+1)}}{\frac{{x}^{2}-{2}^{2}}{3{x}^{2}+9x+6}}n\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\frac{\frac{{x}^{2}(x-2)}{3(x+1)}}{\frac{(x+2)(x-2)}{3{x}^{2}+9x+6}}n\]
Factor out the common term \(3\).
\[\frac{\frac{{x}^{2}(x-2)}{3(x+1)}}{\frac{(x+2)(x-2)}{3({x}^{2}+3x+2)}}n\]
Factor \({x}^{2}+3x+2\).
Ask: Which two numbers add up to \(3\) and multiply to \(2\)?
Rewrite the expression using the above.
\[(x+1)(x+2)\]
\[\frac{\frac{{x}^{2}(x-2)}{3(x+1)}}{\frac{(x+2)(x-2)}{3(x+1)(x+2)}}n\]
Cancel \(x+2\).
\[\frac{\frac{{x}^{2}(x-2)}{3(x+1)}}{\frac{x-2}{3(x+1)}}n\]
Invert and multiply.
\[\frac{{x}^{2}(x-2)}{3(x+1)}\times \frac{3(x+1)}{x-2}n\]
Cancel \(x+1\).
\[\frac{{x}^{2}(x-2)}{3}\times \frac{3}{x-2}n\]
Cancel \(3\).
\[{x}^{2}(x-2)\times \frac{1}{x-2}n\]
Cancel \(x-2\).
\[{x}^{2}n\]
x^2*n
