The term independent of x in expansion of \left(\left(\right. \frac{x + 1}{x^{\frac{2}{3}} \textrm{ }\textrm{ } - \textrm{ }\textrm{ } x^{\frac{1}{3}} + 1} \textrm{ }\textrm{ } - \textrm{ }\textrm{ } \frac{x - 1}{x - x^{\frac{1}{2}}} \left.\right)\right)^{10} is :

Engineering

Mathematics

Introduction to Binomial Theorem

Question

The term independent of x in expansion of ${(\frac{x + 1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{x - 1}{x - x^{\frac{1}{2}}})}^{10}$ is :

210

310

120

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${(\frac{(x^{1 / 3} + 1) (x^{2 / 3} - x^{1 / 3} + 1)}{x^{2 / 3} - x^{1 / 3} + 1} - \frac{(x^{1 / 2} - 1) (x^{1 / 2} + 1)}{x^{1 / 2} (x^{1 / 2} - 1)})}^{10}$

$= {((x^{1 / 3} + 1) - (1 + x^{- 1 / 2}))}^{10}$

$= {(x^{1 / 3} - \frac{1}{x^{1 / 2}})}^{10}$

Let r + 1 is independent term

$∴ T_{r + 1} =^{10} C_{r} {(x^{1 / 3})}^{10 - r} {(\frac{- 1}{x^{1 / 2}})}^{r} = {(- 1)}^{r} \cdot^{10} C_{r} x^{\frac{10 - r}{3} - \frac{r}{2}}$

$∴$ for independent term $\frac{10 - r}{3} - \frac{r}{2} = 0$

$\Rightarrow$ r = 4

$\Rightarrow$ 5^th term is independent

$∴$ T₅ = T_{4 + 1}= (– 1)⁴ · ¹⁰C₄ = 210 Ans.

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