Find $x$ in the binomial expansion $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^x$ if the ratio of the seventh term from the beginning of the binomial expansion to the seventh term from its end is $\dfrac{1}{6}$.
In the expansion of $\left(x+a\right)^n$
$\left(r+1\right)^{th}$ term from the beginning $\;$ $= T_{r+1} = C^n_r \; x^{n-r} \; a^r$
$r^{th}$ term from the end in the expansion of $\left(x+a\right)^n$ is the $\left(n - r + 2\right)^{th}$ term from the beginning
i.e. $\;$ $r^{th}$ term from the end $\;$ $= T_{n-r+2}$
Now, in the expansion of $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^x$
$7^{th}$ term from the beginning $= T_7 = T_{6+1} = C^x_6 \; \left(\sqrt[3]{2}\right)^{x-6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^6$
$7^{th}$ term from the end $= T_{x-7+2} = T_{x-5} = T_{x-6+1} = C^x_{x-6} \; \left(\sqrt[3]{2}\right)^{x-x+6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}$
i.e. $\;$ $T_{x-5} = C^x_{x-6} \; \left(\sqrt[3]{2}\right)^6 \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}$
As per question,
$\dfrac{T_7}{T_{x-5}} = \dfrac{1}{6}$
i.e. $\;$ $\dfrac{C^x_6 \; \left(\sqrt[3]{2}\right)^{x-6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^6}{C^x_{x-6} \; \left(\sqrt[3]{2}\right)^6 \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}} = \dfrac{1}{6}$
Since $\;$ $C^n_r = C^n_{n-r}$ $\implies$ $C^x_6 = c^x_{x-6}$
$\therefore \;$ We have
$\left(\sqrt[3]{2}\right)^{x-12} \times \left(\dfrac{1}{\sqrt[3]{3}}\right)^{12-x} = \dfrac{1}{6}$
i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 2^{\frac{-12}{3}}}{3^{\frac{12}{3}} \times 3^{\frac{-x}{3}}} = \dfrac{1}{6}$
i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 2^{-4}}{3^{4} \times 3^{\frac{-x}{3}}} = \dfrac{1}{6}$
i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 3^{\frac{x}{3}}}{2^4 \times 3^4} = \dfrac{1}{2 \times 3}$
i.e. $\;$ $2^{\frac{x}{3}} \times 3^{\frac{x}{3}} = 2^3 \times 3^3$
$\implies$ $\dfrac{x}{3} = 3$
$\implies$ $x = 9$