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 \(x^2\).
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 \(\eqref{eq:expansion}\) 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 \(\eqref{eq:expansion}\). This comes from \(x^{20}\) in the first bracket and \(x^2\) in the second, giving a coefficient of \(15\).
