In {left( {sqrt[3]{2} + frac{1}{{sqrt[3]{3}}} | vedclass

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(C) The $r^{th}$ term from the beginning is $T_r = ^nC_{r-1} (2^{1/3})^{n-(r-1)} (3^{-1/3})^{r-1}$.For the $7^{th}$ term from the beginning,$r=7$,so $

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$9$

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(C) The $r^{th}$ term from the beginning is $T_r = ^nC_{r-1} (2^{1/3})^{n-(r-1)} (3^{-1/3})^{r-1}$.For the $7^{th}$ term from the beginning,$r=7$,so $T_7 = ^nC_6 (2^{1/3})^{n-6} (3^{-1/3})^6$.The $7^{th}$ term from the end is the $(n-7+1)^{th} = (n-6)^{th}$ term from the beginning.$T_{n-6} = ^nC_{n-7} (2^{1/3})^{n-(n-7)} (3^{-1/3})^{n-7} = ^nC_7 (2^{1/3})^7 (3^{-1/3})^{n-7}$.Given the ratio $\frac{T_7}{T_{n-6}} = \frac{1}{6}$.Since $^nC_6 = ^nC_{n-6}$,the ratio simplifies to $\frac{(2^{1/3})^{n-6} (3^{-1/3})^6}{(2^{1/3})^6 (3^{-1/3})^{n-6}} = \frac{1}{6}$.This simplifies to $\frac{(2^{1/3})^{n-12}}{(3^{-1/3})^{n-12}} = (2^{1/3} \cdot 3^{1/3})^{n-12} = (6^{1/3})^{n-12} = 6^{(n-12)/3} = 6^{-1}$.Equating exponents: $\frac{n-12}{3} = -1$ $\Rightarrow n-12 = -3$ $\Rightarrow n = 9$.

In ${\left( {\sqrt[3]{2} + \frac{1}{{\sqrt[3]{3}}}} \right)^n}$,if the ratio of the $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $\frac{1}{6}$,then $n = $

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