Given [x2/3−x1/3+1x+1−x−x1/2x−1]10
⇒((x1/3+1)−(xx+1))10
⇒(x1/3−x−1/2)10
General term is Tr+1=10Cr(x1/3)10−r(−x−1/2)r
∴310−r−2r=0⇒20−2r−3r=0
⇒r=4
Then, T5=C410=4×3×2×110×9×8×7=210
JEE Main 2021 — Mathematics Algebra
The term independent of x in the expansion of [x2/3−x1/3+1x+1−x−x1/2x−1]10,x=1, is equal to ___.
Held on 18 Mar 2021 · Verified 6 Jul 2026.
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