The term independent of y  in the expansion of (y1/6−y−1/3)9  is

Given expansion (y1/6−y−1/3)9  is 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= 9Cr (y1/6)9−r (−y−1/3)r   Tr+1=  9Cr (−1)r y9−3r6..............(1) y9−3r6 compare with y0 because of independent term⇒y9−3r6=y0 And this will  contain y if 9−3r6=0  ∵ r=3  substitute in Eqn (1) T3+1=  9C3 (−1)3 y9−3×36 T4=  −9×8×73×2×1  y0=−84