In, #(a+b)^n#, the #(r+1)^(th)# term, #T_(r+1)#, is given by,
#(1) : T_(r+1)=""_nC_ra^(n-r)b^r#, where, #r=0,1,2,...,n#.
Our first goal is to find out at which place the const. term occurs. Let it be at #(r+1)^(th)# place, i.e., let it be #T_(r+1)#
In our case, comparing with #(a+b)^n#, we have,
#n=10, a=x^3, and, b=-2/x^2=-2x^-2#. Hence, by #(1)#,
#T_(r+1)=""_10C_r*(x^3)^(10-r)*(-2x^-2)^r#
#=""_10C_r*x^(30-3r)*(-2)^r*x^(-2r)#
#=""_10C_r*(-2)^r*x^(30-3r-2r)#
#=""_10C_r*(-2)^r*x^(30-5r).................(star)#
Let us recall that the const. term has to be free from variable x, and
hence, the index of #x# must be #0#. Accordingly,
#30-5r=0 rArr r=6#
This means that, the const. term is #T_(r+1)=T_7#. By #(star)#, then,
The Const. Term #=T_7=""_10C_6*(-2)^6*x^(30-30)=""_10C_4*(-2)^6#
#=(10xx9xx8xx7)/(1xx2xx3xx4)*64=13440#.
