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As I know $T_{r+{1}}$=$C(n,r)$ I can't able to apply the formula for the Term which not contain $x$,which is applicable for this (${x +\frac{1}{x}})^n$ So, please help me to solve this problem . Any help will be appreciated

Krish
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    Replacing $x$ everywhere with $x^6$ shouldn't change the constant term, but it will get rid of all those fractional exponents, and may let you see some cancellations. – Gerry Myerson Aug 28 '17 at 09:15
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    The first term simplifies to $x^{1/3} + 1$. The second term cannot be simplified unless there is a sign error in the numerator or denominator (but not both). – N. F. Taussig Aug 28 '17 at 09:27

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Assuming the expression is $$\left(\frac{x +1}{x^{2/3}-x^{1/3}+1} -\frac{x-1}{x-x^{1/2}}\right)^{10}$$

The first term can be simplified setting $x^{1/3}=z\to x=z^3$

$\dfrac{x +1}{x^{2/3}-x^{1/3}+1}=\dfrac{z^3+1}{z^2-z+1}=z+1=x^{1/3}+1$

The second term simplifies in a similar way $x^{1/2}=z\to x=z^2$

$\dfrac{x-1}{x-x^{1/2}}=\dfrac{z^2-1}{z^2-z}=1+\dfrac{1}{x^{1/2}}$

The original expression simplifies to $$\left(x^{1/3}+1-1-\dfrac{1}{x^{1/2}}\right)^{10}=\left(\sqrt[3]{x}-\frac{1}{\sqrt{x}}\right)^{10}$$

Thus the terms non containing $x$ are those who simplify the cube and square roots, that is the $+a^6b^4$ term whose coefficient is $+\dbinom{10}{6}=210$

Hope this is useful

Raffaele
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