In the expansion of (x3−1x2)15 , the term void of x  is

Given expansion is (x3−1x2)15 , We have general term in the expansion (x+a)n(∴  Tr+1= nCr xn−r (a)r be the expansion of (x+a)n)   Tr+1= 15Cr (x3)15−r (−1x2)rTr+1= 15Cr (x)45−3r (−1)r (x−2)r         Tr+1= 15Cr x45−5r (−1)r          …. (1)x45−5r   compare with x0 ⇒x45−5r=x0This will not contain x if 45−5r=0∵ r=9  substitute in Eqn (1) then we get∴   Required term =T9+1= 15C9(−1)9                                   =−  15C9=−15C6