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Mathematics Question on Binomial theorem

The term independent of xx in expansion of (x+1x2/3−x1/3+1−x−1x−x1/2)10\bigg(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\bigg)^{10} is

A

4

B

120

C

210

D

310

Answer

210

Explanation

Solution

[x+1x2/3−x1/3+1−(x−1)x−x1/2]10\bigg[ \frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{(x-1)}{x-x^{1/2}}\bigg]^{10}
=[(x1/3)3+13x2/3−x1/3+1−(x)2x(x−1)]10\bigg[ \frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1}-\frac{ \\{ (\sqrt x)^2 \\}}{\sqrt x (\sqrt x-1)}\bigg]^{10}
=[(x1/3+1)(x2/3+1−x1/3)x2/3−x1/3+1−(x)2−1x(x−1)]10\bigg[ \frac{(x^{1/3}+1)(x^{2/3}+1-x^{1/3})}{x^{2/3-x^{1/3}+1}}-\frac{ \\{(\sqrt x)^2-1 \\}}{\sqrt x(\sqrt x-1)}\bigg]^{10}
=[(x1/3+1)−(x+1)x]10   =(x1/3−x−1/2)10=\bigg[ (x^{1/3}+1)-\frac{(\sqrt x+1)}{\sqrt x}\bigg]^{10} \, \, \, = (x^{1/3}-x^{-1/2})^{10}
∴\therefore The general term is
Tr+1= 10Cr(x1/3)10−r(−x−1/2)r= 10Cr(−1)rx10−r3−r2T_{r+1}= \, ^{10}C_r(x^{1/3})^{10-r} (-x^{-1/2})^r = \, {10}C_r(-1) ^r x^{\frac{10-r}{3}-\frac{r}{2}}
For independent of x , put
10−r3−r2=0   ⇒   20−2r−3r=0\frac{10-r}{3}-\frac{r}{2}=0 \, \, \, \Rightarrow \, \, \, 20-2r-3r=0
⇒            20=5r  ⇒  20−2r−3r=0\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, 20=5r \, \, \Rightarrow \, \, 20-2r-3r=0
∴  T5= 10C4=10× 9× 8×74×3×2×1=210\therefore \, \, T_5= \, ^{10}C_4=\frac{10\times \, 9 \times \, 8 \times 7}{4 \times 3 \times 2 \times 1}=210