Question

The term independent of $x$ in the expansion of ${(\sqrt[6]{6x}-\frac{1}{\sqrt[3]{3x}})}^9$ is equal to:

• A. ${-}^9{C}_3$
• B. ${}^9{C}_3$
• C. ${}^9{C}_2$
• D. none of the above

Answer 1

Answer: (A)

${-}^9{C}_3$

Given

${(\sqrt[6]{6x}-\frac{1}{\sqrt[3]{3x}})}^9$

General term of given expansion is

$T_{r+1}={}^9{C}_r\left(x{)}^{\frac{9-r}{6}}\frac{1}{(x{)}^{\frac{r}{3}}}\right(-1{)}^r$

$T_{r+1}={}^9{C}_r\left(x{)}^{\frac{9-3r}{6}}\right(-1{)}^r-----\left(1\right)$

Comparing terms having power $x^0$ with $T_{r+1}={}^9{C}_r\left(x{)}^{\frac{9-3r}{6}}\right(-1{)}^r$ we get

$\frac{9-3r}{6}=0\Rightarrow r=3$

Hence coefficient of term having term $x^0$ is ${}^9{C}_3(-1{)}^3=-{}^9{C}_3$