Problem source

\begin{questionparts} 
\item Find the coefficient of $x^{6}$ in
\[(1-2x+3x^{2}-4x^{3}+5x^{4})^{3}.\]
You should set out your working clearly.
 
\item By considering the binomial expansions of $(1+x)^{-2}$ and $(1+x)^{-6}$, or otherwise, find the
coefficient of $x^{6}$ in
\[(1-2x+3x^{2}-4x^{3}+5x^{4}-6x^{5}+7x^{6})^{3}.\]
\end{questionparts}

Solution source

\begin{questionparts}
\item We can obtain a $6$ from $4+2+0, 4+1+1, 3+3+0, 3+2+1, 2+2+2$.

So $x^6$ from $4,2,0$ can happen in $6$ ways and gets us a coefficient of $1 \cdot 3 \cdot 5$.
$x^6$ from $4,1,1$ can happen in $3$ ways and gets us a coefficient of $5 \cdot (-2) \cdot (-2)$.
$x^6$ from $3,3,0$ can happen in $3$ ways and gets us a coefficient of $(-4) \cdot (-4) \cdot 1$.
$x^6$ from $3,2,1$ can happen in $6$ ways and gets us a coefficient of $(-4) \cdot 3 \cdot (-2)$.
$x^6$ from $2,2,2$ can happen in $1$ ways and gets us a coefficient of $3 \cdot 3 \cdot 3$.

This leaves us with a total coefficient of:

$6 \cdot 15 + 3 \cdot 20 + 3 \cdot 16 + 6 \cdot 24 + 1 \cdot 27 = 369$

\item \begin{align*}
(1+x)^{-2} &= 1 + (-2)x+\frac{(-2)\cdot(-3)}{2!} x^2 + \frac{(-2)(-3)(-4)}{3!}x^3 + \cdots \\
&= 1 -2x+3x^2-4x^3+5x^4+\cdots \\
\end{align*}

The coefficient of $x^6$ in the expansion of $(1+x)^{-6}$ will be $\frac{(-6)(-7)(-8)(-9)(-10)(-11)}{6!} = \frac{11!}{6!5!} = 462$.

The coefficient of $x^6$ in the expansion of $(1 -2x+3x^2-4x^3+5x^4+\cdots)^3$ will be the same as the coefficient of $x^6$ in the expansion of $(1 -2x+3x^2-4x^3+5x^4-6x^5+7x^6)^3$, ie it will be $462$
\end{questionparts}