If the coefficient of x ^ 15 in the expansion of (a x ^ 3 +...

$( r + 1{)}^{th}$ term in the expansion of ${\left(a{x}^{3} + \dfrac{1}{bx\dfrac{1}{3}}\right)}^{15}$ is

${T}_{r + 1} = Cr 15 {\left(a{x}^{3}\right)}^{15 - r}{\left(\dfrac{1}{b{x}^{1 / 3}}\right)}^{r}$

$⇒ {T}_{r + 1} = Cr 15 {\left(a\right)}^{15 - r}{\left(x\right)}^{45 - \dfrac{10r}{3}}{\left(\dfrac{1}{b}\right)}^{r}$

For the coefficient of ${x}^{15}$ in ${\left(a{x}^{3} + \dfrac{1}{b{x}^{1 / 3}}\right)}^{15}$:

$45 - \dfrac{10r}{3} = 15$

$⇒ 30 = \dfrac{10r}{3}$

$⇒ r = 9$

Coefficient of ${x}^{15} = C9 15 {a}^{6}{b}^{- 9}$

$( r + 1{)}^{th}$ term in the expansion of ${\left(a{x}^{1 / 3} - \dfrac{1}{b{x}^{3}}\right)}^{15}$ is:

${T}_{r + 1} = Cr 15 {\left(a{x}^{1 / 3}\right)}^{15 - r}{\left(- \dfrac{1}{b{x}^{3}}\right)}^{r}$

For the coefficient of ${x}^{- 15}$ in ${\left(a{x}^{1 / 3} - \dfrac{1}{b{x}^{3}}\right)}^{15}$ is:

$5 - \dfrac{r}{3} - 3r = - 15$

$⇒ \dfrac{10r}{3} = 20$

$⇒ r = 6$

Coefficient of ${x}^{- 15}$ $= C6 15 {a}^{9} \times {b}^{- 6}$

Hence,

$C9 15 {a}^{6}{b}^{- 9} = C6 15 {a}^{9} \times {b}^{- 6}$

$⇒ \dfrac{{a}^{9}}{{b}^{6}} = \dfrac{{a}^{6}}{{b}^{9}}$

$⇒ {a}^{3}{b}^{3} = 1 ⇒ ab = 1$