Finding coefficients in expansions Show that the coefficient of $x^{−12}$ in the expansion of
$$\left(x^4−\frac{1}{x^2}\right)^5\left(x−\frac{1}{x}\right)^6$$
is $−15$, and calculate the coefficient of $x^2$. Hence, or otherwise, calculate the coefficients of $x^4$ and $x^{38}$ in the expansion of $$(x^2−1)^{11}(x^4+x^2+1)^5.$$
The first part of this exercise is solved easily considering "partitions", i.e. what can be added to get the exponents and turns out to be $-15$ and $215$ respectively. I was able to solve the second part of this exercise easily using multinomial theorem. However, this exercise is given before multinomial theorem is introduced, and I would really like to know what the "Hence" was gunning for.  Once I factor out to get$$x^{21}\left(x-\frac{1}{x}\right)^{11}\left(x^2+\frac{1}{x^2}+1\right)^5,$$I can see similarities with the first question. I can even deduce that since $4-21=-17$, $38-21=17$, and inspecting signs, that the coefficients are equal in magnitude and opposite in sign. Please explain how I can use the first part of the question to proceed, rather than going "otherwise"/multinomial theorem.

\begin{align*}
(x^2 - 1)^{11}(x^4 + x^2 + 1)^5 
&=(x^2 - 1)^{5}(x^4 + x^2 + 1)^5(x^2 - 1)^{6} \\
&=(x^6 - 1)^5(x^2 - 1)^6 \\
&=x^{16}\left(x^4 - \frac{1}{x^2}\right)^5\left(x - \frac{1}{x}\right)^6
\end{align*}
Thus, the coefficient of $x^4$ in the new equation is the same as the coefficient of $x^{-12}$ in the original equation, which is $-15$.
And like you mentioned, the coefficient of $x^4$ and $x^{42 - 4} = x^{38}$ should be the same but opposite in sign. So the coefficient of $x^{38}$ is $15$.
