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

Q: Why do we simplify the expression before expanding? Can't we just use the binomial theorem on the original mess?

While technically possible, expanding a binomial with such complex fractions would lead to a nightmare of terms. Simplifying first using algebraic identities is the 'JEE way'—it turns a 30-minute calculation into a 3-minute one. Always look for simplification opportunities first!

Q: What if the value of $r$ we get is not an integer? Does that mean there is no term independent of $x$?

Exactly! If $r$ is not an integer between $0$ and $n$, it implies that no term in the expansion is independent of $x$. In this problem, we got $r=4$, which is a valid integer, so we are safe.

Q: Why is the negative sign in the second term so important?

It's crucial because of the $(-1)^r$ factor. If $r$ were odd, the term would be negative. Even though it didn't change the sign here (since $r=4$ is even), ignoring it is a common trap that leads to incorrect coefficients in many problems.

Question

The term independent of $x$ in the expansion of $\left( \frac{(x+1)}{(x^{2/3}+1-x^{1/3})} - \frac{(x+1)}{(x-x^{1/2})} \right)^{10}, x > 1$ is:

Options:

(A) 210

(B) 150

(C) 240

(D) 120

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, traveler of the JEE landscape. Today, we face a problem that, at first glance, looks like a monster. We are presented with a binomial expression raised to the power of ten, containing fractions that seem designed to intimidate.

Our mission is to find the term independent of $x$. This means we are hunting for the constant term, the one where the power of $x$ vanishes into $x^0$.

### Phase 1: The Power of Substitution

Look at the expression: 
$$\left( \frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}} ight)^{10}$$

It is a mess, but notice the fractional powers. They are screaming for a substitution. Let us set $x^{1/3} = a$, which implies $x = a^3$.

The first fraction becomes:
$$\frac{a^3+1}{a^2-a+1}$$

Now, do you recognize the sum of cubes identity? Since $a^3+1 = (a+1)(a^2-a+1)$, the denominator is exactly the second factor. They cancel out, leaving us with just $a+1$, or $x^{1/3} + 1$.

Now, for the second fraction: $\frac{x-1}{x-x^{1/2}}$. Let $x^{1/2} = b$, so $x = b^2$.

The numerator is $b^2-1$, which is a difference of squares: $(b-1)(b+1)$. The denominator is $b^2-b = b(b-1)$.

Canceling the $(b-1)$ terms, we get:
$$\frac{b+1}{b} = 1 + \frac{1}{b} = 1 + x^{-1/2}$$

### Phase 2: The Beautiful Collapse

Now, let us bring these simplified pieces back into the original expression. We have $(x^{1/3} + 1) - (1 + x^{-1/2})$.

Look at that! The ones cancel out perfectly, as $1 - 1 = 0$. The entire expression collapses into:
$$(x^{1/3} - x^{-1/2})^{10}$$

What was once a terrifying, multi-fractional nightmare is now a simple binomial. This is the beauty of algebra; it rewards those who look for the underlying structure.

### Phase 3: The Binomial Hunt

Now that we have $(x^{1/3} - x^{-1/2})^{10}$, we use the general term formula:
$$T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r$$

We need to isolate the variable $x$. Let us group the powers:
$$x^{\frac{10-r}{3}} \cdot (-1)^r \cdot x^{-\frac{r}{2}}$$

Combining the exponents, we get $x^{\frac{10-r}{3} - \frac{r}{2}}$. For the term to be independent of $x$, the exponent must be zero.

So, we solve:
$$\frac{10-r}{3} - \frac{r}{2} = 0$$

Multiplying by 6, we get $2(10-r) - 3r = 0$, which simplifies to $20 - 2r - 3r = 0$, or $5r = 20$. Thus, **$r=4$**.

### Phase 4: The Final Victory

We have found our $r$. Now, substitute $r=4$ back into the constant part of our general term:
$$\binom{10}{4} (-1)^4$$

Since $(-1)^4 = 1$, the term is simply $\binom{10}{4}$. Calculating this:
$$\frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$$

We have conquered the beast. The term independent of $x$ is **210**.

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

Select Answer:

Visualized Solution

Expression Analysis

Strategy: Simplifying the First Fraction

Applying Algebraic Identity

Simplifying the Second Fraction

Canceling Terms in Second Fraction

Combining the Simplified Terms

The General Term Formula

Substituting into the General Term

Combining the Powers of x

Setting the Condition for Independence

Solving for r

Calculating the Coefficient

Final Computation

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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 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 ___

The term independent of $x$ in the expansion of $(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}$ , where $x \neq = 0, 1$ is equal to

The term independent of x in the expansion of $(\frac{1}{60} - \frac{x ^{8}}{81}) (2 x^{2} - \frac{3}{x ^{2}})^{6}$ is equal to :

The term independent of $x$ in the expression of $(1 - x^{2} + 3 x^{3}) (\frac{5}{2} x^{3} - \frac{1}{5 x ^{2}})^{11}$ , $x \neq = 0$ is

The maximum value of the term independent of 't' in the expansion of $(t x^{1/5} + \frac{( 1 - x ) ^{1/10}}{t})^{10}$ where $x \in (0, 1)$ is:

If the term independent of $x$ in the expansion of $(a x^{2} + \frac{1}{2 x ^{3}})^{10}$ is 105 , then $a^{2}$ is equal to :

Let the coefficients of third, fourth and fifth terms in the expansion of $(x + \frac{a}{x ^{2}})^{n}, x \neq = 0$ , be in the ratio $12 : 8 : 3$ . Then the term independent of $x$ in the expansion, is equal to ____.

If the greatest value of the term independent of $x$ in the expansion of $(x sin α + a \frac{c o s α}{x})^{10}$ is $\frac{10 !}{( 5 ! ) ^{2}}$ , then the value of $a$ is equal to:

If the term independent of $x$ in the expansion of $(\frac{3}{2} x^{2} - \frac{1}{3 x})^{9}$ is $k$ , then $18 k$ is equal to :

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+1−x1/3)(x+1)​−(x−x1/2)(x+1)​)10,x>1 is: