Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& (1+\mathrm{x}{)}^{\mathrm{p}}(1-\mathrm{x}{)}^{\mathrm{q}}=\left({}^{\mathrm{p}}{\mathrm{C}}_0+{}^{\mathrm{p}}{\mathrm{C}}_1\mathrm{x}+{}^{\mathrm{p}}{\mathrm{C}}_2{\mathrm{x}}^2+\dots \right)\left({}^{\mathrm{q}}{\mathrm{C}}_0-{}^{\mathrm{q}}{\mathrm{C}}_1\mathrm{x}+{}^{\mathrm{q}}{\mathrm{C}}_2{\mathrm{x}}^2+\dots \right)\\ & \text{ coff of }\mathrm{x}\equiv {}^{\mathrm{p}}{\mathrm{C}}_0{}^{\mathrm{q}}{\mathrm{C}}_1+{}^{\mathrm{p}}{\mathrm{C}}_1{}^{\mathrm{q}}{\mathrm{C}}_0=1\\ & \mathrm{p}-\mathrm{q}=1\\ & \text{ coff of }{\mathrm{x}}^2\equiv {}^{\mathrm{p}}{\mathrm{C}}_0{}^{\mathrm{q}}{\mathrm{C}}_2-{}^{\mathrm{p}}{\mathrm{C}}_1{}^{\mathrm{q}}{\mathrm{C}}_1+{}^{\mathrm{p}}{\mathrm{C}}_2{}^{\mathrm{q}}{\mathrm{C}}_0=-2\\ & \frac{\mathrm{q}(\mathrm{q}-1)}{2}-\mathrm{pq}+\frac{\mathrm{p}(\mathrm{p}-1)}{2}=-2\\ & {\mathrm{q}}^2-\mathrm{q}-2\mathrm{pq}+{\mathrm{p}}^2-\mathrm{p}=-4\\ & 1-(\mathrm{p}+\mathrm{q})=-4\\ & \mathrm{p}+\mathrm{q}=5\\ & \mathrm{p}=3\\ & \mathrm{q}=2\\ & \text{ so }{\mathrm{p}}^2+{\mathrm{q}}^2=13\end{array}$