Given that $(1+(2/3)x)^n \cdot (3+*n*x)^2 = 9+84x+\dotsb$ find $*n*$.
I started off by figuring out the second half of it-- obviously $3+nx$ to the power of 2 is equal to $n^2x^2 + 6nx + 9$. Then I tried to multiply each individual quantity by the first group( $(1+(2/3)x)^n$ ). I then tried to divide $84x$ by $6nx$ and I realized I couldn't do that, and basically from there on out I was completely lost. For context, this is an International Baccalaureate Higher Level Math question.
Hint:
The $x$ term of $(1+\frac{2}{3}x)^n$ is $$\binom{n}{1}1^{n-1}\left( \frac{2}{3}x\right)=\frac{2n}{3}x$$.
The $x$ term of $ (3+nx)^2$ is $nx$.
So, the $x$ term of the product is: $$3\frac{2n}{3}x+nx=3nx$$.