If the coefficients of x ^ 7 in ( x ^ 2 + (1)/(bx)) ^ 11 and x ^ -...

Coefficient of ${x}^{7}$ in ${\left({x}^{2} + \dfrac{1}{bx}\right)}^{11}$

$= Cr 11 {\left({x}^{2}\right)}^{11 - r} \cdot {\left(\dfrac{1}{bx}\right)}^{r}$

$= Cr 11 {x}^{22 - 3r} \cdot \dfrac{1}{{b}^{r}}$

So, $22 - 3r = 7$

$⇒ r = 5$

So, the term is $C5 11 \cdot \dfrac{1}{{b}^{5}} \cdot {x}^{7}$

Coefficient of ${x}^{- 7}$ in ${\left(x - \dfrac{1}{b{x}^{2}}\right)}^{11}$

$= Cr 11 {\left(x\right)}^{11 - r} \cdot {\left(- \dfrac{1}{b{x}^{2}}\right)}^{r}$

$= Cr 11 {x}^{11 - 3r} \cdot \dfrac{{\left(- 1\right)}^{r}}{{b}^{r}}$

So, $11 - 3r = - 7 ∴ r = 6$

So, the term is $C6 11 \cdot \dfrac{1}{{b}^{6}}{x}^{- 7}$

Hence, $C5 11 \cdot \dfrac{1}{{b}^{5}} = C6 11 \cdot \dfrac{1}{{b}^{6}}$

Since $b \ne 0 ∴ b = 1$