If the coefficient of x⁷ in ${\left({\mathrm{ax}}^2+\frac{1}{\mathrm{bx}}\right)}^{11}$ equals to the coefficient of x^-7 in ${\left({\mathrm{ax}}^2-\frac{1}{\mathrm{bx}}\right)}^{11}$ then a and b satisfy___

$\mathrm{In}{\left({\mathrm{ax}}^2+\frac{1}{\mathrm{bx}}\right)}^{11};{\mathrm{T}}_{\mathrm{r}+1}={11}_{{\mathrm{c}}_{\mathrm{r}}}\frac{{\mathrm{a}}^{11-\mathrm{r}}}{{\mathrm{b}}^{\mathrm{r}}}{\mathrm{x}}^{22-3\mathrm{r}}\text{ for }{\mathrm{x}}^7$ coefficient ;
$22-3\mathrm{r}=7\Rightarrow 3\mathrm{r}=15\Rightarrow \mathrm{r}=5$
 $\therefore$coefficient of  ${\mathrm{x}}^7={11}_{{\mathrm{c}}_5}\frac{{\mathrm{a}}^6}{{\mathrm{b}}^5}\to \left(1\right)$
In ${\left(\mathrm{ax}-\frac{1}{{\mathrm{bx}}^2}\right)}^{11};{\mathrm{T}}_{\mathrm{r}+1}=(-1{)}^{\mathrm{r}}{11}_{{\mathrm{c}}_{\mathrm{r}}}\frac{{\mathrm{a}}^{11-\mathrm{r}}}{{\mathrm{b}}^{\mathrm{r}}}{\mathrm{r}}^{11-3\mathrm{r}}$
For x^-7 coefficient; 
$11-3\mathrm{r}=-7\Rightarrow 3\mathrm{r}=18\Rightarrow \mathrm{r}=6$
$\therefore$coefficient of ${\mathrm{x}}^{-7}={11}_{{\mathrm{c}}_6}\frac{{\mathrm{a}}^5}{{\mathrm{b}}^6}\to \left(2\right)$
$\because \left(1\right)=\left(2\right)\Rightarrow \mathrm{ab}=1$