<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] Sure it does work. As you let $P(x)=ax^3+bx^2+cx$, we have $P(x)-P(x-1)=3ax^2-(3a-2b)x+a-b+c$
Now just compare the coefficients. That is, $3a=1$, $3a-2b=0$, $a-b+c=0$ which after solving gives $a=\frac13 , b=\frac12 , c=\frac16$.
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