<s>[INST] How to find $a$, $b$, $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$ I'm trying to find $a$, $b$ and $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$.
After expanding the binomial in $P(x-1)$, I end up getting
$3ax^2-3ax+2bx+a-b=x^2$. What next? Using $3a = 1$ doesn't work.
 [/INST] Alternatively. We see that $p(0)=0$.
If $x=1$ we get $$1 = p(1)-p(0) = a+b+c \implies \boxed{a+b+c=1}$$
If $x=2$ we get $$4 = p(2)-p(1) = 7a+3b+c \implies \boxed{7a+3b+c=4}$$
If $x=-1$ we get $$1 = p(0)-p(-1) = a-b+c \implies \boxed{a-b+c=1}$$
Now solve this system...
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