Animated Solution for Mathematics - Binomial Theorem: The term independent of $x$ in expansion of $(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}{)}^{10}$ is

The expression provided is:

At first glance, this appears to be a daunting expansion. However, in JEE Advanced mathematics, complexity is often a mask for underlying algebraic identities. We will simplify the expression by breaking it into two distinct parts.

A Hidden Identity

Let us isolate the first fraction:

If we let $a=x^{1/3}$, then the denominator becomes $a^2-a+1$. Since $x=a^3$, the numerator is $a^3+1^3$.

Applying the sum of cubes identity, $a^3+b^3=(a+b)(a^2-ab+b^2)$, we rewrite the numerator as:

Substituting this back into the fraction, the term $(x^{2/3}-x^{1/3}+1)$ cancels out perfectly. We are left with:

The Difference of Squares

Now, let us examine the second fraction:

Let $u=x^{1/2}$, which implies $x=u^2$. The numerator becomes $u^2-1^2$, which factors as $(u-1)(u+1)$. The denominator becomes $u^2-u$, which factors as $u(u-1)$.

The expression becomes:

Canceling the $(x^{1/2}-1)$ term, we are left with:

We now combine the simplified fractions into the original expression:

The constants $1$ and $-1$ cancel each other out, leaving us with a much simpler binomial:

To find the term independent of $x$, we use the general term formula $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$, where $n=10$, $a=x^{1/3}$, and $b=-x^{-1/2}$.

The general term is:

Isolating the powers of $x$, we get:

For the term to be independent of $x$, the exponent must be zero:

Multiplying by 6, we obtain $2(10-r)-3r=0$, which simplifies to $20-5r=0$, yielding $r=4$.

Substituting $r=4$ into the coefficient part of the general term:

The final value of the term independent of $x$ is $210$.