Let $P(x)$ be the polynomial $x^3+Ax^2+Bx+C$ for some constants $A$, $B$, and $C$. There exists constants $D$ and $E$ such that for all $x$, $P(x+1)=x^3+Dx^2+54x+37$ and $P(x+2)=x^3+26x^2+Ex+115$. Compute the ordered triple $(A,B,C)$.
Try to substitute values of $x$ into $P(x+1)$ and $P(x+2)$ to make your life easier.
On
$P(x+1)=x^3+Dx^2+54x+37$
$P(x+1)=(x+1)^3+A(x+1)^2+B(x+1)+C=x^3+3x^2+3x+1+Ax^2+2Ax+A+Bx+B+C=x^3+x^2(A+3)+x(3+A+B)+(1+A+B+C)$
Comparing this to our values:
$$1+A+B+C=37\implies A+B+C=36$$
$$A+3=D$$
$$3+A+B=54 \implies A+B=51$$
Substituting $A+B$, $51+C=36$, and thus $C=-15$
Continue with the same process, incorporating:
$P(x+2)=x^3+26x^2+Ex+115$ and
$P(x+2)=(x+2)^3+A(x+2)^2+B(x+2)+C$, then grouping.
In $P(x+2)$ we notice that we have that nasty $Ex$. We want to get rid of that term with a substitution for $x$. The way we do this is to substitute $0$. This gives us that $P(2) = 0 + 0 + 0 + 115 = 115$. By substituting $1$ into $P(x+1)$, you get $P(2)=92 + D$. We can solve the equation $92+D=115$. This gives us $D=23$. Now, we can substitute $(x+1)$ for all $x$ in $P(x+1)$. This gives us another polynomial (which I don't want to write out cuz LaTeX). This gives us the final result of $x^3+20x^3+11x+5$ for $P(x)$. Our ordered pair is $(A,B,C) = (20,11,5)$.