The term independent of $x$ in the expansion of $\left( {\frac{1}{{60}} - \frac{{{x^8}}}{{81}}} \right).{\left( {2{x^2} - \frac{3}{{{x^2}}}} \right)^6}$ is equal to

  • [JEE MAIN 2019]
  • A

    $36$

  • B

    $-36$

  • C

    $-108$

  • D

    $-72$

Similar Questions

If the constant term in the binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell}}\right)^9$ is $-84$ and the Coefficient of $x^{-3 \ell}$ is $2^\alpha \beta$, where $\beta < 0$ is an odd number, Then $|\alpha \ell-\beta|$ is equal to

  • [JEE MAIN 2023]

If the co-efficient of $x^9$ in $\left(\alpha x^3+\frac{1}{\beta x}\right)^{11}$ and the co-efficient of $x^{-9}$ in $\left(\alpha x-\frac{1}{\beta x^3}\right)^{11}$ are equal, then $(\alpha \beta)^2$ is equal to $.............$.

  • [JEE MAIN 2023]

If the coefficients of $x^{7}$ in $\left(x^{2}+\frac{1}{b x}\right)^{11}$ and $x^{-7}$ in $\left(x-\frac{1}{b x^{2}}\right)^{11}, b \neq 0$, are equal, then the value of $b$ is equal to:

  • [JEE MAIN 2021]

Coefficient of ${x^2}$ in the expansion of ${\left( {x - \frac{1}{{2x}}} \right)^8}$ is

The middle term in the expansion of ${\left( {x + \frac{1}{x}} \right)^{10}}$ is