Recall that:
[tex]a^2-b^2=(a+b)(a-b).[/tex]
Then:
[tex]x^2-9=(x+3)(x-3).[/tex]
Also, notice that:
[tex]3x+9=3x+3*3=3(x+3).[/tex]
Therefore:
[tex]\frac{3x+9}{x^3}*\frac{2x}{x^2-9}=\frac{3(x+3)}{x^3}*\frac{2x}{(x+3)(x-3)}.[/tex]
Assuming that x≠-3 and x≠0 we can rewrite the above result as follows:
[tex]\frac{3}{x^2}*\frac{2}{(x-3)}.[/tex]
Therefore:
[tex]\frac{3x+9}{x^3}*\frac{2x}{x^2-9}=\frac{6}{x^3-3x^2}.[/tex]
Answer: Option E. No mistakes.
