The term independent of x in the expansion of 160−x881⋅2x2−3x26is equal to

Let a binomial 2x2−3x26, it's (r+1) th term =Tr+1=6Cr2x26−r−3x2r=6Cr(−3)r(2)6−rx12−2r−2r  ---(i)=6Cr(−3)r(2)6−rx12−4r       Now, the term independent of  x in the expansion of 160−x8812x2−3x26= the term independent of x in the expansion of 1602x2−3x26 the term independent of x in the expansion of −x8812x2−3x26= 6C360(−3)3(2)6−3x12−4(3)+−1816C5(−3)5(2)6−5x12−4(5)x8=13(−3)323+35×2(6)81=36−72=−36