<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] 
For the sake of a different method:

$$
\begin{cases}
P(x)-P(x-1)=x^2  \\
P'(x)-P'(x-1)=2x  \\
P''(x)-P''(x-1)=2 
\end{cases} 
$$
or writing derivatives explicitly:
$$
\begin{cases}
a\big(x^3-(x-1)^3\big)+b\big(x^2-(x-1)^2\big)+c\big(x-(x-1)\big)=x^2~~~~~~~(1) \\
3a\big(x^2-(x-1)^2\big)-2b\big(x-(x-1)\big)=2x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2) \\
6a\big(x-(x-1)\big)=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (3)
\end{cases}
$$
Beginning from the end, it's very easy to solve for $a,b,c$:

*

*First, find $\boxed{a=\frac{1}{3}}$ from the $(3)$.

*Second, plug $a$ into $(2)$ and find $\boxed{b=\frac{1}{2}}$.

*Third, plug $a$ and $b$ into $(1)$ and find $\boxed{c=\frac{1}{6}}$
Each step yields a trivial linear equation.
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