In the expansion of (12x1/3+x−1/5)8, the term independent of x is
T6
T6
Suppose the (r+1)th term in the given expansion is independent of x.
Then, we have:
Tr+1=8Cr(12x13)8−r(x−15)r
=8Cr128−rx3−r5
For this term to be independent of x, we must have:
8−r3−r5=0
⇒40−5r−3r=0
⇒r=5
Hence, the required term is the 6th term, i.e.T6.