Mathematics - Binomial Theorem Question with Solution | TestHub

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

$T_{r+1}={}^{15}C_r{\left(a{x}^3\right)}^{15-r}{\left(\frac{1}{b{x}^{1/3}}\right)}^r$

$\Rightarrow T_{r+1}={}^{15}C_r{\left(a\right)}^{15-r}{\left(x\right)}^{45-\frac{10r}{3}}{\left(\frac{1}{b}\right)}^r$

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

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

$\Rightarrow 30=\frac{10r}{3}$

$\Rightarrow r=9$

Coefficient of $x^{15}={}^{15}C_9{a}^6{b}^{-9}$

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

$T_{r+1}={}^{15}C_r{\left(a{x}^{1/3}\right)}^{15-r}{\left(-\frac{1}{b{x}^3}\right)}^r$

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

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

$\Rightarrow \frac{10r}{3}=20$

$\Rightarrow r=6$

Coefficient of $x^{-15}$ $={}^{15}\mathrm{C}_6{\mathrm{a}}^9\times {\mathrm{b}}^{-6}$

Hence,

${}^{15}C_9{a}^6{b}^{-9}={}^{15}\mathrm{C}_6{\mathrm{a}}^9\times b^{-6}$

$\Rightarrow \frac{a^9}{b^6}=\frac{a^6}{b^9}$

$\Rightarrow a^3{b}^3=1\Rightarrow ab=1$