<s>[INST] Distributing multiplication of rational functions I am having trouble distributing with fractions. This
$$ \left(\frac{1}{(x + 3)} + \frac{(x + 3)}{(x - 3)}\right)\, (9 - x^2) = -\frac{(x^2-3)}{9-x^2}(9-x^2) $$
has the answer $\left\{\left[x = - \frac{9}{7}\right]\right\}$.
I start solving by cancelling $(9-x^2)$ on the right hand side, and multiplying $(9-x^2)$ with the terms on the left hand side by foiling (multiplying term by term). So the first step is
$$9\times\frac{1}{x+3}+\frac{9 (x+3)}{x-3}-x^2\times\frac{1}{x+3}-\frac{x^2 (x+3) }{x-3}=-x^2-3$$
This is apparently wrong because it has another answer, $\left\{\left[x = - \frac{3}{7}\right]\right\}$ What am I doing wrong?
 [/INST] $\left(\frac{1}{x+3} + \frac{x+3}{x-3} \right) ( 9 -x^2) -x^2 -3$ 
It will be 
$\frac{9-x^2}{x+3} + \frac{(x+3)(9-x^2)}{x-3} = -x^2 - 3 $ 
after that 
$\frac{(9-x^2)(x-3)}{(x+3)(x-3)} + \frac{(x+3)(x+3)(9-x^2)}{(x-3)(x+3)} = -x^2 -3 $ 
Note that $9-x^2 = -(x^2 - 9) = -(x+3)(x-3)$, i.e., you can cancel it with the denominator 
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