Need help getting started with a complicated limits problem So I've got this thing 
$\lim_{x \rightarrow 3} \frac{\sqrt[3]{x^2+6x}-3}{x-3}$
I tried rationalizing but then it ended up in a huge mess and I didn't get the correct answer. I don't know if there's something else to do before resorting to rationalization. I just need something to get started. 
Aside from that I know that something on the numerator will cancel out because if I evaluate the limit it ends up as $\frac{0}{0}$

Hint
$$\lim_{x \to 3} \frac{\sqrt[3]{x^2+6x}-3}{x-3}=\lim_{x\to 3} \frac{\left(\sqrt[3]{x^2+6x}-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}{\left(x-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}$$
$$=\lim_{x\to 3} \frac{x^2+6x-27}{\left(x-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}=\lim_{x\to 3} \frac{x+9}{\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9}$$
