Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively. Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively.Find $(a-b).$

I tried to factorize $(1+x+2x^2+3x^3)$ and $(1+x+2x^2+3x^3+4x^4)$ into product of two binomials,but i could not.And i do not have any other method to solve it.
 A: Let $p(x)=1+x+2x^2+3x^3$ and $q(x)=1+x+2x^2+3x^3+4x^4$. Then 
$$(q(x))^4-(p(x))^4=(q(x)-p(x))A(x)$$
for some polynomial $A(x)$. But $q(x)-p(x)=4x^4$, so $(q(x))^4-(p(x))^4$ is divisible by $x^4$. It follows that all coefficients of $x^k$ for $k\le 3$ are $0$.
A: The  problem is a way of making the point that changing the terms of degree $k$ and higher in the argument can only change the degree $k$-or-higher terms in the result, when applying polynomial or power-series operations (such as taking the 4th power of the argument).  If you are performing a calculation to get the answer, something is wrong.
A: Here is a slightly different variation of the theme. If we introduce the polynomial
\begin{align*}
p(x)=1+x+2x^2+3x^3
\end{align*}
and write $[x^n]$ to denote the coefficient of $x^n$ of a polynomial,

we observe
  \begin{align*}
&[x^3]\left(p(x)\right)^4-[x^3]\left(p(x)+4x^4\right)^4\\
&\quad=[x^3]\left(p(x)\right)^4-[x^3]\left(p(x)\right)^4\tag{1}\\
&\quad=0
\end{align*}
  In (1) we use the fact that multiplication of $4x^4$ with $p(x)$ gives always terms with power of $x$ greater or equal to $4$ and so there is no contribution to $[x^3]$.

