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.
 A: 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 $a=\frac13 , b=\frac12 , c=\frac16$.
A: 
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.
A: 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...
