<s>[INST] What is the coefficient of the $x^3$ term in the expansion of $(x^2+x-5)^7$ (See details)? I fail to see a simple way to answer this.
As such, this is my long winded approach:
Using the multinomial theorem,
$$(x_1 + x_2  + \cdots + x_m)^n 
 = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m}
  \prod_{1\le t\le m}x_{t}^{k_{t}},$$
we have the specific parameters $m=3$, $n=7$, $x_1=x^2$, $x_2=x$, and $x_3=-5$.
Via the theorem,
$$(x^2+x-5)^7=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}\prod_{1 \le t \le 3}x_{t}^{k_t}=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}x^{2k_1}x^{k_2}(-5)^{k_3}.$$
The coefficient of the $x^3$ term is the summation of the multinomial coefficient multiplied by the $(-5)^{k_3}$ factor evaluated at all the solutions of the equation $2k_1+k_2=3$ where $0\le k_1\le 7$ and $0 \le k_2 \le 7$.
Those values are $(k_1,k_2)=\{(1,1),(0,3)\}.$ Given that $k_1+k_2+k_3=7$, $k_3$ are respectively $5$ and $4$.
Hence, the coefficient of the $x^3$ is
$${7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4.$$
Given that the definition of the multinomial coefficient is
$${n \choose k_1,k_2,\ldots ,k_m}=\frac{n!}{k_1!k_2!\cdots k_m!},$$
$$
\begin{align}
{7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4&=\frac{7!(-5)^5}{1!1!5!}+\frac{7!(-5)^4}{0!3!4!}\\
&=7\cdot 6(-5)^5+\frac{7\cdot 6\cdot 5(-5)^4}{3!}\\
&=-109375
\end{align}$$
This could be atrociously wrong. Either way, I am desperate for a much simpler process. This is ridiculous to do in a timed testing environment without the formulas given.
I would like to see a very simple but also very general way of arriving at the correct answer (preferably without college methods, but I am open to any methods).
What says you, Math.SE?
 [/INST] This is pretty much the same as joriki's approach, but the emphasis here is on a slightly different idea. Write
$$
[x^2+x+(-5)]^7=\sum_{k=0}^7{7\choose k}(x^2+x)^k(-5)^{7-k}\,.
$$
Now note that for $k\ge 4$, the power $(x^2+x)^k$ starts with $x^4$ or higher power, and for $k<2$, it ends with $x^2$ or lower power. Thus only 2 terms ($k=2$ and $k=3$) have $x^3$ in them. Looking at what the corresponding coefficients are, we get
$$
{7\choose 3}(-5)^4+{7\choose 2}\cdot 2\cdot (-5)^5
$$
The rest of the computation is the same as in joriki's answer.
</s>