$$\begin{gathered} \text{ If the coefficient of }{\mathrm{x}}^7\text{ in }{\left(\mathrm{ax}-\frac{1}{{\mathrm{bx}}^2}\right)}^{13}\text{ and the coefficient of }{\mathrm{x}}^{-5}\text{ in }{\left(\mathrm{ax}+\frac{1}{{\mathrm{bx}}^2}\right)}^{13}\text{ are equal, then }{\mathrm{a}}^4{\mathrm{b}}^4\text{ is equal} \\ \text{to: } \end{gathered}$$

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{expansion} \mathrm{of} {\left(\mathrm{ax}-\frac{1}{{\mathrm{bx}}^2}\right)}^{13} \\ {\mathrm{T}}_{\mathrm{r}+1}{=}^{13}{\mathrm{C}}_{\mathrm{r}}(a\mathrm{x}{)}^{13-\mathrm{r}}{\left(-\frac{1}{{\mathrm{bx}}^2}\right)}^{\mathrm{r}} \\ 13-3\mathrm{r}=7\Rightarrow \mathrm{r}=2 \end{gathered}$$

$$\begin{gathered} \text{ Similarly coeff. of }{x}^{-5}\text{ in }{\left(ax+\frac{1}{b{x}^2}\right)}^{13} \\ {\mathrm{T}}_{\mathrm{r}+1}{=}^{13}{\mathrm{C}}_{\mathrm{r}}(a\mathrm{x}{)}^{13-\mathrm{r}}{\left(\frac{1}{{\mathrm{bx}}^2}\right)}^{\mathrm{r}} \\ 13-3\mathrm{r}=-5 \\ for {\text{r=6 is }}^{13}{C}_6\frac{a^7}{b^6} \end{gathered}$$

${ }^{13}{C}_2\frac{a^{11}}{b^2}{=}^{13}{C}_6\frac{a^7}{b^6}$

$\Rightarrow a^4{b}^4=\frac{{ }^{13}{C}_6}{{ }^{13}{C}_2}=22$