If the co-efficient of x ^ 9 in (α x ^ 3 + (1)/( )) ^ 11 and the...

Given,

The coefficient of ${x}^{9}$ in ${\left(\alpha{x}^{3} + \dfrac{1}{\betax}\right)}^{11}$and the coefficient of ${x}^{- 9}$ in ${\left(\alphax - \dfrac{1}{\beta{x}^{3}}\right)}^{11}$ are equal,

Now the ${r}^{th}$ term of ${\left(\alpha{x}^{3} + \dfrac{1}{\betax}\right)}^{11}$ is given by $= Cr 11 \cdot \dfrac{{\alpha}^{11 - r}}{{\beta}^{r}} \times {x}^{3\left(11 - r\right) - r}$

So, for ${x}^{9}$, $33 - 4r = 9 ⇒ r = 6$

Hence, coefficient of ${x}^{9}$ in ${\left(\alpha{x}^{3} + \dfrac{1}{\betax}\right)}^{11} = C6 11 \cdot \dfrac{{\alpha}^{5}}{{\beta}^{6}}$

Similarly, for coefficient of ${x}^{- 9}$ in ${\left(\alphax - \dfrac{1}{\beta{x}^{3}}\right)}^{11} = - C5 11 \cdot \dfrac{{\alpha}^{6}}{{\beta}^{5}}$

Now given both coefficient are equal,

So, $\dfrac{11}{{C}_{6}} \cdot \dfrac{{\alpha}^{5}}{{\beta}^{6}} = - \dfrac{11}{{C}_{5}} \cdot \dfrac{{\alpha}^{6}}{{\beta}^{5}}$

$⇒ \dfrac{1}{\beta} = - \alpha$

$⇒ \alpha\beta = - 1$

$⇒ ( \alpha\beta{)}^{2} = 1$