In the expansion \( \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n} \) if the ratio between the seventh term from the beginning to the seventh term from the end \( 1: 6 \), then \( n \) equals \( \begin{array}{llll}\text { (a) } 7 & \text { (b) } 8 & \text { (c) } 9 & \text { (d) } 10\end{array} \)

Let's denote the terms in the binomial expansion. The general term \( T_k \) in the expansion of \( \left(a + b\right)^n \) is given by: \[ T_k = \binom{n}{k-1} a^{n-k+1} b^{k-1} \] In our case, \( a = \sqrt[3]{2} \) and \( b = \frac{1}{\sqrt[3]{3}} \). Thus, we have: \[ T_k = \binom{n}{k-1} \left(\sqrt[3]{2}\right)^{n-k+1} \left(\frac{1}{\sqrt[3]{3}}\right)^{k-1} \] For the seventh term from the beginning, which is \( T_7 \): \[ T_7 = \binom{n}{6} \left(\sqrt[3]{2}\right)^{n-6} \left(\frac{1}{\sqrt[3]{3}}\right)^{6} \] For the seventh term from the end, which is \( T_{n-6} \): \[ T_{n-6} = \binom{n}{n-6} \left(\sqrt[3]{2}\right)^{6} \left(\frac{1}{\sqrt[3]{3}}\right)^{n-6} \] The ratio of these two terms is given as \( 1:6 \): \[ \frac{T_7}{T_{n-6}} = \frac{\binom{n}{6} \left(\sqrt[3]{2}\right)^{n-6} \left(\frac{1}{\sqrt[3]{3}}\right)^{6}}{\binom{n}{n-6} \left(\sqrt[3]{2}\right)^{6} \left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}} = \frac{1}{6} \] Since \( \binom{n}{6} = \binom{n}{n-6} \), we can simplify the equation to: \[ \frac{\binom{n}{6}}{\binom{n}{6}} \cdot \frac{\left(\sqrt[3]{2}\right)^{n-6}}{\left(\sqrt[3]{2}\right)^{6}} \cdot \frac{\left(\frac{1}{\sqrt[3]{3}}\right)^{6}}{\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}} = \frac{\left(\sqrt[3]{2}\right)^{n-6}}{\left(\sqrt[3]{2}\right)^{6}} \cdot \left(\frac{1}{\sqrt[3]{3}}\right)^{6-n+6} \] Let’s simplify this expression: \[ \frac{\left(\sqrt[3]{2}\right)^{n-6}}{\left(\sqrt[3]{2}\right)^{6}} = \left(\sqrt[3]{2}\right)^{n-12} \] \[ \frac{\left(\frac{1}{\sqrt[3]{3}}\right)^{6-n+6}}{\left(\frac{1}{\sqrt[3]{3}}\right)^{0}}= \left(3^{1/3}\right)^{n-12} = \left(\sqrt[3]{3}\right)^{n-12} \] Putting this together, we get: \[ \left(\sqrt[3]{2}\right)^{n-12} \left(\sqrt[3]{3}\right)^{n-12} = \frac{1}{6} \] Now, let’s denote \( \sqrt[3]{2} = 2^{1/3} \) and \( \sqrt[3]{3} = 3^{1/3} \): \[ \left(\frac{2}{3}\right)^{(n-12)/3} = \frac{1}{6} \] Taking the cube of both sides: \[ \frac{2}{3}^{n-12} = \frac{1}{216} \] This implies: \[ \frac{2}{3}^{n-12} = \frac{1}{6^3} = \left(\frac{1}{2}\cdot\frac{1}{3}\right)^3 \] Since \( \frac{2}{3}^{n-12} = (2^{-3}) \cdot (3^{-3}) \