2016
Problem - 3117
Let $P(x)$ be the polynomial $x^3 + Ax^2 +Bx+C$ for some constants $A, B,$ and $C$. There exists constant $D$ and $E$ such that for all $x$, $P(x+1)=x^3 + Dx^2 + 54x +37$ and $P(x+2)=x^3 + 26x + Ex+115$. Compute the ordered triple $(A, B, C)$.
Setting $x=1$ to $P(x+1)$ gives $$P(2)=1+D+54+37=D+92$$
Meanwhile, letting $x=0$ in $P(x+2)$ yields $$P(2)=115$$
Therefore, we find $D=115-92=23$ or $$P(x+1)=x^3 + 23x^2 + 54x+37$$
Let $y=x+1$ or $x=y-1$. We have
$$P(y)=(y-1)^3+23(y-1)^2 + 54(y-1)+37=y^3+20y^2+11y+5$$
Therefore, we conclude
$$(A,B,C)=\boxed{(20, 11, 5)}$$