$$f(x^3) - f(x^3 - 2) = (f(x))^2 + 12$$
Given the functional equation above, I am trying to find the value of $f(3)$.
I do not remember the exact statement of the problem precisely, so I am not sure whether an initial value of the form $f(a) = p$ was provided.
I started off by finding the degree of $f(x)$ on both sides, which must be equal. If the degree of $f(x)$ is $n$, then the right side clearly has degree $2n$, while the left side seems to have a degree of the form $3(n-1)$.
Equating this, I got that $f(x)$ is a polynomial of degree $3$.
Now I am stuck — I thought of setting $f(x) = a x^3 + b x^2 + cx + d$ but I don't believe that will help much.
EDIT: Ok I did that, and substituted back into the functional equation to get:
$$a x^9-a \left(x^3-2\right)^3+b x^6-b \left(x^3-2\right)^2+c x^3-c \left(x^3-2\right)=\left(a x^3+b x^2+c x+d\right)^2+12$$
Ew. I guess I could expand and equate coefficients on both sides (and indeed, churching through that with Mathematica gives $6x^3 - 6$), but seeing as this is an AIME-esque problem, there should probably be an easier way. Is there one?