If the coefficient of $x^{15}$ in the expansion of $\left(\mathrm{a} x^{3}+\frac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15}$ is equal to the coefficient of $x^{-15}$ in the expansion of $\left(a x^{1 / 3}-\frac{1}{b x^{3}}\right)^{15}$, where $a$ and $b$ are positive real numbers, then for each such ordered pair $(\mathrm{a}, \mathrm{b})$ :

<p>For $\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)$</p> <p>$${T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}$$</p> <p>$\therefore$ ${x^{15}} \to 3(15 - r) - {r \over 3} = 15$</p> <p>$\Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9$</p> <p>Similarly, for ${\left( {a{x^{{1 \over 3}}} - {1 \over {b{x^3}}}} \right)^{15}}$</p> <p>$${T_{r + 1}} = {}^{15}{C_r}{\left( {a{x^{{1 \over 3}}}} \right)^{15 - r}}{\left( { - {1 \over {b{x^3}}}} \right)^2}$$</p> <p>$\therefore$ For ${x^{ - 15}} \to {{15 - r} \over 3} - 3r = - 15 \Rightarrow r = 6$</p> <p>$\therefore$ $${}^{15}{C_9}{{{a^6}} \over {{b^9}}} = {}^{15}{C_6}{{{a^9}} \over {{b^6}}} \Rightarrow ab = 1$$</p>