Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$.

Question

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: $$=\boxed{\dfrac{2x}{3x+1}}$$

Answer 1

HINT: Pull out everything that you can: $6x^3-2x^2=2x^2(3x-1)$, and $9x^3-x=x(9x^2-1)$. Then notice that $9x^2=(3x)^2$, so that $9x^2-1=(3x)^2-1^2=(3x-1)(3x+1)$. Finally, do the cancellations that are now apparent.

Answer 2

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, $$\frac{\left(6x^3-2x^2\right)}{\left(9x^3-x\right)}=\frac{2 x^2 (3x-1 )}{x (3x-1) (3x+1)}=\hspace{2pt}\ldots\hspace{2pt}\Large{?}$$

Answer 3

$$\dfrac{(6x^3-2x^2)}{x(9x^3-x)} = \dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} $$

Cancel common factor of $x$ in numerator and denominator gives us:

$$\dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} = \frac{2x(3x-1)}{\left[9x^2 - 1\right]}$$

Now we have a difference of squares in the denominator, and can factor it:

$$\frac{2x(3x-1)}{\color{blue}{\bf \left[9x^2 - 1\right]}}= \frac{2x(3x-1)}{\color{blue}{\bf \left[(3x)^2 - 1\right]}} = \frac{2x(3x - 1)}{(3x - 1)(3x + 1)}$$

Now, cancel like terms: Note that $\color{blue}{(3x - 1)}$ is a factor in the numerator and in the denominator, so we proceed to cancel:

$$ \frac{2x\color{blue}{\bf (3x - 1)}}{\color{blue}{\bf(3x - 1)}(3x + 1)} = \frac{ 2x}{3x + 1}$$