Checking Independence for $\left\{\dfrac{1}{x^2+x-6},\dfrac{1}{x^2-5x+6},\dfrac{1}{x^2-9}\right\}$? 
Is this subset of $V=\mathbb{F[0,1]}$ linearly independent, where the subset of $V$ is 
  $\left\{\dfrac{1}{x^2+x-6},\dfrac{1}{x^2-5x+6},\dfrac{1}{x^2-9}\right\}$?

My work: 
If this subset is independent, then $\alpha=\beta=\gamma=0$ must be the only solution for $\alpha\dfrac{1}{x^2+x-6}+\beta\dfrac{1}{x^2-5x+6}+\gamma\dfrac{1}{x^2-9}=0\\ \implies\dfrac{\alpha}{x^2+x-6}+\dfrac{\beta}{x^2-5x+6}+\dfrac{\gamma}{x^2-9}=0\\\implies \dfrac{\alpha}{(x+3)(x-2)}+\dfrac{\beta}{(x-2)(x-3)}+\dfrac{\gamma}{(x+3)(x-3)}=0\\$
I'm note sure how to proceed from this step (assuming what I've done so far is correct). A hint would be appreciated. Thanks.

Hint: checking that this set is linearly independent is equivalent to checking if the set 
$$\{x-3, x-2, x+3\}$$ is linearly independent.
