The term independent of x in the expansion of (1/60 - (x⁸)/81).(2x² - 3/(x²))⁶ is equal to:

Binomial Theorem and Its Simple Applications: JEE Main 2019 Mathematics question

JEE Main 2019 · 12 Apr, Evening shift · Q2914

JEE Main 201912 Apr, Evening shiftMathematicsMediumBinomial Theorem and Its Simple Applications131× since 2002

The term independent of x in the expansion of

\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}

is equal to :

JEE Main 2019 · 12 Apr, Evening shift · Q2914

JEE Main 201912 Apr, Evening shiftMathematicsMediumBinomial Theorem and Its Simple Applications131× since 2002

The term independent of x in the expansion of

\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}

is equal to :