<s>[INST] Functional equation $f(x^3) - f(x^3 - 2) = (f(x))^2 + 12$ $$f(x^3) - f(x^3 - 2) = (f(x))^2 + 12$$
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?
 [/INST] Note that the left side depends only on $x^3$, so the right side should also.  Thus we look for a function of the form $f(x) = g(x^3)$.  As you say, $f$ should have degree $3$, so $g$ should have degree $1$.  If $f(x) = a x^3 + b$, then taking $x = 0$ we get $b - (-8a + b) = b^2 + 12$, i.e. 
$8 a = b^2 + 12$.  Substitute $a= (b^2+12)/8$  in to the equation, 
expand, and subtract right from left: I get
$$ \left(-\frac{b^4}{64} + \frac{3}{8} b^2 + \frac{27}{4}\right) x^6
+ \left(-\frac{b^3}{4} - \frac{3}{2} b^2 - 3 b - 18 \right) x^3 = 0
$$
so both the coefficients of $x^6$ and $x^3$ must be $0$.  Factoring the first, which is a quadratic in $b^2$:
$$ -\frac{b^4}{64} + \frac{3}{8} b^2 + \frac{27}{4} = - \frac{1}{64} (b^2 - 36)(b^2 + 12) = 0 $$
so if we're looking for real solutions, $b = \pm 6$.  Of those only $b = -6$ makes the other coefficient $0$.  Thus the only polynomial solution
with real coefficients is $b = -6$, $a = 6$: $f(x) = 6 x^3 - 6$.
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