Question

In the binomial expansion of ${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^x$ is ,the ratio at the $7$ term from the beginning and $7$ term to end is $1:6$ then find ${}^{\prime }{x}^{\prime }$

• A. $9$
• B. $6$
• C. $12$
• D. $Noneofthese$

Answer 1

The correct option is A $9$

We have,

${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^x$

Since,

$T_7{=}^x{C}_6\times {\left(2^{\frac{1}{3}}\right)}^{x-6}\times {\left(\frac{1}{3^{\frac{1}{3}}}\right)}^6$

$7thtermsfromendin{\left(3\sqrt{2}+\frac{1}{3\sqrt{3}}\right)}^x$

${=}^x{C}_6\times {\left(\frac{1}{3^{\frac{2}{3}}}\right)}^{x-6}\times {\left(2^{\frac{1}{3}}\right)}^6$

According to the question,

$\frac{C_6\times {\left(2^{\frac{1}{3}}\right)}^{x-6}\times {\left(\frac{1}{3^{\frac{1}{3}}}\right)}^6}{C_6\times {\left(3^{\frac{1}{3}}\right)}^{x-6}\times {\left(2^{\frac{1}{3}}\right)}^6}=\frac{1}{6}{2}^{\frac{x-12}{3}}\times 3^{\frac{x-12}{3}}=\frac{1}{6}{6}^{\frac{x-12}{3}}=\frac{1}{6}x-12=-3x=9$

Hence, this is the answer.