MathematicsMedium64×since 2002Q2914The term independent of x in the expansion of (160−x881).(2x2−3x2)6\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}(601−81x8).(2x2−x23)6 is equal to :A36B- 108C- 36D- 72Check answerSkip