Recursive Polynomial Problem Source: Purple Comet Competition 2017 High School Question 12
Let $P(x)$ be a polynomial satisfying $P(x+1) + P(x-1) = x^3$ for all real numbers $x$. Find the value of $P(12)$.
I am pretty sure one is supposed to write down several "formulas", and then cancel everything out to leave $P(12)$ behind, but I cannot get the answer and would like some help in solving this problem. What is the correct and easiest way to do this problem?
 A: Clearly the polynomial is cubic, and coefficient of $x^3$ is $1/2$.
So $p(x)=(1/2)x^3+ax^2+bx+c$
Now we have $P(2)=1-c$ and $P(-2)=-1-c$
so $4+4a+2b=1-c$ and
$-4+4a-2b=-1-c$
Adding the two eqn's we have $a=-c/4$ and $b=-3/2$.
Now the constant term of $P(x+1)+P(x-1)$ is $1/2+a+b+c-1/2+a-b+c=2a+2c=0$
So, $a=-c$, hence $a=c=0$
So our polynomial is $(x^3-3x)/2$.
So $p(12)=846$.
A: Hint: $\;P(0)=0$ since the RHS has no free term is odd. Then, using $P(x+1)=x^3-P(x-1)\,$:
$$
P(12)=11^3 - P(10) = 11^3-9^3+ P(8)= \cdots=11^3-9^3+7^3-5^3+3^3-1^3+P(0)
$$

[ EDIT ]  As pointed out in a comment, why $\,P(0)=0\,$ needs some additional elaboration.

Main point is that the RHS of the given identity is an odd function. Writing the equalities for $\,x\,$ and $\,-x\,$, then adding up together, gives:
$$
P(x+1)+P(x-1)+P(-x+1)+P(-x-1) = x^3 + (-x)^3=0
$$
With $\,Q(x)=P(x)+P(-x)\,$ the above can be rewritten as:
$$
Q(x+1)+Q(x-1) = 0 \;\;\iff\;\; Q(x+2)+Q(x)=0
$$
That implies that for each root $\,a\,$ of $\,Q(x)\,$, $a+2$ is also a root. Since a non-zero polynomial cannot have infinitely many roots, it follows that $\,Q(x) \equiv 0\,$. This in turn implies that $\,P(x)+P(-x) \equiv 0\,$ and substituting $\,x=0\,$ gives $\,P(0)=0\,$.
