The term independent of x in the expansion of [(x+1)/(x^(2/3)-x^(1/3)+1) - (x-1)/(x-x^(1/2))]^10, x

Q: Wait, why can't I just expand the binomial directly? It's only to the power of 10.

That is a classic trap! Expanding a binomial with complex fractional terms directly would lead to a massive, unmanageable mess of terms. In JEE problems, there is almost always a hidden simplification. By simplifying the base first, you turn a 'nightmare' problem into a standard, solvable binomial expansion.

Q: What if the value of $r$ I get isn't an integer? Does that mean I made a mistake?

Exactly! In binomial expansion, $r$ must be a non-negative integer between $0$ and $n$. If your calculation for $r$ results in a fraction or a negative number, it implies that there is no term independent of $x$ in that specific expansion. However, in this problem, the math works out perfectly to $r=4$.

Q: Do I need to worry about the negative sign in the second term?

Absolutely. When you write the general term, the second term is $-x^{-1/2}$. When you raise this to the power of $r$, you must include the $(-1)^r$ factor. If $r$ were odd, the term would be negative; since $r=4$ is even, it becomes positive. Always keep track of that sign!

Question

The term independent of $x$ in the expansion of $[\frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}}]^{10}, x 
eq 1$, is equal to ___

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating expression:

$$ \left[ \frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}} ight]^{10} $$

Many students would immediately try to expand this using the binomial theorem, but that is a path to frustration. In the world of JEE Advanced, we don't fight the math; we outsmart it.

### Simplifying the Interior

Our first step is to simplify the interior. Look at the first fraction: $\frac{x+1}{x^{2/3}-x^{1/3}+1}$.

The denominator screams for the sum of cubes identity. If we set $a = x^{1/3}$ and $b = 1$, then $a^3 + b^3 = (x^{1/3})^3 + 1^3 = x+1$.

The identity $a^3 + b^3 = (a+b)(a^2-ab+b^2)$ allows us to rewrite the numerator as $(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)$. When we divide this by the denominator, the complex part cancels out perfectly, leaving us with just $x^{1/3} + 1$.

Now, look at the second fraction: $\frac{x-1}{x-x^{1/2}}$. This is a difference of squares in disguise.

We can write $x-1$ as $(\sqrt{x}-1)(\sqrt{x}+1)$ and the denominator as $\sqrt{x}(\sqrt{x}-1)$. Again, the $(\sqrt{x}-1)$ terms cancel, leaving us with $\frac{\sqrt{x}+1}{\sqrt{x}}$, which simplifies to $1 + x^{-1/2}$.

### The Binomial Lens

Now, let's bring these simplified pieces back into our original expression. We have $[(x^{1/3} + 1) - (1 + x^{-1/2})]^{10}$.

Notice the magic? The $+1$ and $-1$ cancel each other out completely! We are left with $(x^{1/3} - x^{-1/2})^{10}$.

This is the moment where the 'nightmare' becomes a beautiful, elegant binomial. To find the term independent of $x$, we invoke the general term formula:

$$ T_{r+1} = {}^{n}C_r a^{n-r} b^r $$

Here, $n=10$, $a=x^{1/3}$, and $b=-x^{-1/2}$. Substituting these, we get:

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

### The Hunt for Zero

We are not just looking for any term; we are hunting for the term independent of $x$. This means the power of $x$ must be exactly zero.

Let's group the powers of $x$:

$$ x^{\frac{10-r}{3}} \cdot x^{-\frac{r}{2}} = x^{\frac{10-r}{3} - \frac{r}{2}} $$

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

$$ \frac{10-r}{3} - \frac{r}{2} = 0 $$

Multiplying by $6$ to clear the fractions, we get $2(10-r) - 3r = 0$, which simplifies to $20 - 2r - 3r = 0$, or $20 - 5r = 0$. Solving this, we find $r=4$.

### Final Calculation

Finally, we calculate the coefficient by substituting $r=4$ back into the constant part of our general term: ${}^{10}C_4 (-1)^4$.

Since $(-1)^4 = 1$, we just need to calculate:

$$ {}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210 $$

The beast is tamed. The **answer is 210**.

Animated Solution for Mathematics - Binomial Theorem: The term independent of $x$ in the expansion of $[\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}}]^{10}, x \neq = 1$ , is equal to ___

Enter Numerical Value:

Visualized Solution

Analyze the Expression

Simplify First Fraction: Identity

Simplify First Fraction: Execution

Simplify Second Fraction: Identity

Simplify Second Fraction: Execution

Combine Simplified Terms

General Term Formula

Substitute into General Term

Simplify Exponents

Condition for Independent Term

Solve for $r$

Calculate Final Coefficient

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Simplifying the Interior

The Binomial Lens

The Hunt for Zero

Final Calculation

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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 ___