Can we find the coefficients of $x^{−12}$ and $x^2$ in this expansion?

Can we find the coefficients of $x^{−12}$ and $x^2$ in this expansion?

Question

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 $(x^2 − 1)^{11}(x^4 + x^2 + 1)^5.$

Suggestion

Could we expand each of the brackets?

What contribution do we need from each of the bracketed terms to get $x^{-12}$ or $x^2$ terms?

How are the two parts related? Could we rewrite this new expression to make it look like the first part?

What happens if we mulitply $(x^4+x^2+1)$ by $(x^2-1)$?

Solution

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 $x^2$.

Using the binomial expansion on each bracket, we have

$$\begin{align} \left(x^4-\frac{1}{x^2}\right)^5 &\left(x-\frac{1}{x}\right)^6\not ag \\ =\left(x^{20}-5x^{14}+10x^8-10x^2+5x^{-4}-x^{-10}\right) &\left(x^{6}-6x^{4}+15x^2-20+15x^{-2}-6x^{-4}+x^{-6}\right). \end{align}$$

The only way we can get an $x^{-12}$ term on multiplying out is from the $x^{-10}$ term in the first bracket multiplied by the $x^{-2}$ term in the second bracket.

Thus the coefficient of $x^{-12}$ in the full expansion is $(-1)\times15 = -15$.

There are three ways we can get an $x^{2}$ term on multiplying out: $x^{8} \times x^{-6}$, $x^{2} \times x^{0}$ and $x^{-4} \times x^{6}$.

Thus the coefficient of $x^{2}$ in the full expansion is $10 + 200 + 5 = 215$.

Hence, or otherwise, calculate the coefficients of $x^4$ and $x^{38}$ in the expansion of $(x^2 − 1)^{11}(x^4 + x^2 + 1)^5.$

We can note first that

$$\begin{align*} \left(x^4-\frac{1}{x^2}\right)^5\left(x-\frac{1}{x}\right)^6 &= \left(\frac{x^6-1}{x^2}\right)^5\left(\frac{x^2-1}{x}\right)^6\\ &= \dfrac{(x^6-1)^5(x^2-1)^6}{x^{16}}. \end{align*}$$

Secondly, we can note that $(x^4+x^2+1)(x^2-1) = x^6 - 1.$

Thus

$$\begin{align*} (x^2-1)^{11}(x^4+x^2+1)^5 &= (x^2-1)^{6}(x^2-1)^{5}(x^4+x^2+1)^5\\ &=(x^2-1)^{6}(x^6-1)^{5}\\ &=\dfrac{x^{16}(x^2-1)^{6}(x^6-1)^{5}}{x^{16}}\\ &= x^{16}\left(x^4-\frac{1}{x^2}\right)^5\left(x-\frac{1}{x}\right)^6. \end{align*}$$

So to find the coefficient of $x^4$ in $(x^2-1)^{11}(x^4+x^2+1)^5$ is to find the coefficient of $x^{-12}$ in the expansion $(expansion)$ of $\left(x^4-\dfrac{1}{x^2}\right)^5\left(x-\dfrac{1}{x}\right)^6$.

We already know this to be $-15$.

And to find the coefficient of $x^{38}$ in $(x^2-1)^{11}(x^4+x^2+1)^5$ is to find the coefficient of $x^{22}$ in the expansion $(expansion)$. This comes from $x^{20}$ in the first bracket and $x^2$ in the second, giving a coefficient of $15$.