<s>[INST] Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$. Tried this one a couple of times but can't seem to figure it out.
I am trying to simplify the expression:
$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$
So my attempt at this is:
$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\div\bigg(\dfrac{9x^2}{x^2}-\dfrac{1}{x^2}\bigg)$$
$$=\bigg(\dfrac{6x-2}{x}\bigg)\div\bigg(\dfrac{9x^2-1}{x^2}\bigg)$$
$$=\dfrac{6x-2}{x}\cdot\dfrac{x^2}{9x^2-1}$$
$$=\dfrac{(6x-2)(x^2)}{(x)(9x^2-1)}$$
$$=\dfrac{6x^3-2x^2}{9x^3-x}$$
This is the part that I get stuck at. I can't decide what to factor out:
$$=\dfrac{x(6x^3-2x^2)}{x(9x^3-x)}$$
$$=\dfrac{(6x^2-2x)}{(9x^2-1)}$$
Edit, missed a difference of squares:
$$=\dfrac{2x^2(6x^3-2x^2)}{x(9x^3-x)}$$
$$=\dfrac{2x^2(3x-1)}{x(3x-1)(3x+1)}$$
Giving a final answer of:
$$=\boxed{\dfrac{2x}{3x+1}}$$
 [/INST] You can factor a $2x^2$ out of the top to get $6x^3-2x^2=2 x^2 (3x-1 )$.  After this, you can factor an $x$ out of the bottom to get $(9x^2-1)$, which then factors into $(3x-1) (3x+1)$ as it is a difference of squares.  So, putting that all together,
$$\frac{\left(6x^3-2x^2\right)}{\left(9x^3-x\right)}=\frac{2 x^2 (3x-1 )}{x (3x-1) (3x+1)}=\hspace{2pt}\ldots\hspace{2pt}\Large{?}$$
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