If the coefficients x⁷ in ${\left({\mathrm{ax}}^2+\frac{1}{\mathrm{bx}}\right)}^{11}$and coefficient of x^-7in ${\left(\mathrm{ax}-\frac{1}{{\mathrm{bx}}^2}\right)}^{11}$are equal, then the value of ab is _____.

Let ${\mathrm{x}}^7\text{ occurs in }{\mathrm{T}}_{\mathrm{r}+1}$term, then 
${\mathrm{T}}_{\mathrm{r}+1}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\left({\mathrm{ax}}^2\right)}^{\mathrm{n}-\mathrm{r}}{\left(\frac{1}{\mathrm{bx}}\right)}^{\mathrm{r}}{=}^{11}{\mathrm{C}}_{\mathrm{r}}\frac{{\mathrm{a}}^{11-\mathrm{r}}}{{\mathrm{b}}^{\mathrm{r}}}\cdot {\mathrm{x}}^{22-2\mathrm{r}-\mathrm{r}}$
For ${\mathrm{x}}^7,22-3\mathrm{r}=7\text{ or }\mathrm{r}=5$
Hence, coefficients of ${\mathrm{x}}^7{\text{ is }}^{11}{\mathrm{C}}_5\frac{{\mathrm{a}}^6}{{\mathrm{b}}^5}$.
Let ${\mathrm{x}}^{-7}\text{ occur in }{\mathrm{T}}_{\mathrm{r}+1}$term then 
${\mathrm{T}}_{\mathrm{r}+1}{=}^{11}{\mathrm{C}}_{\mathrm{r}}(\mathrm{ax}{)}^{11-{\left(-\frac{1}{{\mathrm{bx}}^2}\right)}^{\mathrm{r}}}{=}^{11}{\mathrm{C}}_{\mathrm{r}}\frac{{\mathrm{a}}^{11-\mathrm{r}}}{(-\mathrm{b}{)}^{\mathrm{r}}}{\mathrm{x}}^{11-3\mathrm{r}}$
For ${\mathrm{x}}^{-7}\Rightarrow 11-3\mathrm{r}=-7\Rightarrow \mathrm{r}=6$
Hence, coefficient of  ${\mathrm{x}}^{-7}{\text{ is }}^{11}{\mathrm{C}}_6\frac{{\mathrm{a}}^5}{{\mathrm{b}}^6}$
Now, ${ }^{11}{\mathrm{C}}_5\frac{{\mathrm{a}}^5}{{\mathrm{b}}^6}{=}^{11}{\mathrm{C}}_6\frac{{\mathrm{a}}^6}{{\mathrm{b}}^5}$
$$\begin{gathered} \begin{array}{l}{\Rightarrow }^{11}{\mathrm{C}}_5\mathrm{a}{=}^{11}{\mathrm{C}}_6\frac{{\mathrm{a}}^5}{{\mathrm{b}}^6}\\ {\Rightarrow }^{11}{\mathrm{C}}_5\mathrm{a}{=}^{11}{\mathrm{C}}_{11-6}\frac{1}{\mathrm{b}}\\ {\Rightarrow }^{11}{\mathrm{C}}_5\mathrm{a}{=}^{11}{\mathrm{C}}_5\frac{1}{\mathrm{b}}\end{array} \\ \Rightarrow \mathrm{ab}=1 \end{gathered}$$