In ${\left( {\sqrt[3]{2} + \frac{1}{{\sqrt[3]{3}}}} \right)^n}$ if the ratio of ${7^{th}}$ term from the beginning to the ${7^{th}}$ term from the end is $\frac{1}{6}$, then $n = $
In ${\left( {\sqrt[3]{2} + \frac{1}{{\sqrt[3]{3}}}} \right)^n}$ if the ratio of ${7^{th}}$ term from the beginning to the ${7^{th}}$ term from the end is $\frac{1}{6}$, then $n = $
- A
$7$
- B
$8$
- C
$9$
- D
None of these
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