Problems

Solution: The powers of \(x\) in the first bracket will be \(x^{20}, x^{14}, \cdots, x^{-10}\). The powers of \(x\) in the second bracket will be \(x^6, x^4, \cdots, x^{-6}\). Therefore we can achieve \(x^{-12}\) in only one way: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{-10} & x^{-2} & \binom{5}{5}(-1)^5 = -1 & \binom{6}{4}(-1)^4 = 15& -15 \\ \end{array} We can achieve \(x^2\) as follows: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{-4} & x^{6} & \binom{5}{4}(-1)^4 = 5 & \binom{6}{0}(-1)^0 = 1& 5 \\ x^{2} & x^{0} & \binom{5}{3}(-1)^3 = -10 & \binom{6}{3}(-1)^3 = -20 & 200 \\ x^{8} & x^{-6} & \binom{5}{2}(-1)^2 = 10 & \binom{6}{6}(-1)^6 = 1 & 10 \end{array} Therefore the coefficient is \(215\) \((x^2-1)(x^4+x^2+1) = x^6-1\), therefore \begin{align*} (x^2-1)^{11}(x^4+x^2+1)^5 &= (x^2-1)^6(x^6-1)^5 \\ &= x^6\left(x-\frac1x\right)^6(x^6-1)^6 \\ &= x^6\left(x-\frac1x\right)^6\left(x^2\left(x^4-\frac{1}{x^2}\right)\right)^5 \\ &= x^6\left(x-\frac1x\right)^6x^{10}\left(x^4-\frac{1}{x^2}\right)^5 \\ &= x^{16}\left(x-\frac1x\right)^6\left(x^4-\frac{1}{x^2}\right)^6 \\ \end{align*} Therefore the coefficient of \(x^4\) is the coefficient of \(x^{4-16} = x^{-12}\) in our original expression, ie \(-15\). Similarly, the coefficient of \(x^{38}\) is the coefficient of \(x^{38-16} = x^{22}\), which can only be achieved in one way: \begin{array}{c|c|c|c|c} 1\text{st bracket} & 2\text{nd bracket} & 1\text{st coef} & 2\text{nd coef} & \text{prod} \\ \hline x^{20} & x^{2} & \binom{5}{0}(-1)^0 = 1 & \binom{6}{2}(-1)^2 = 15& 15 \\ \end{array} Therefore the coefficient is \(15\)