The term independent of x in the expansion of (1/60 - x⁸/81)(2x² - 3/x²)⁶ is equal to : (A) 36 (B) -

Q: Why can't I just find the general term for the whole expression at once?

The binomial theorem applies to expressions of the form $(a+b)^n$. Here, you have a product of a binomial and a polynomial. You must distribute the terms first, which creates two separate cases to solve.

Q: I got $r=3$ and $r=5$. Are these the only possible values?

Yes, because $r$ must be an integer between $0$ and $n=6$. If you had calculated an $r$ outside this range, that term would simply not exist.

Q: Why did the negative sign in $-\frac{3}{x^2}$ matter so much?

It changes the sign of the coefficient depending on whether $r$ is even or odd. Always treat the term as $(a + b)$ where $b = -3/x^2$. Ignoring the sign is the most common way to lose marks in these problems.

Question

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

Options:

(A) 36

(B) -108

(C) -72

(D) -36

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the term independent of $x$ in the expression $(\frac{1}{60} - \frac{x^8}{81})(2x^2 - \frac{3}{x^2})^6$, we must identify the components where the net power of $x$ is zero.

The expression is a product of a binomial and a polynomial. We will analyze the expansion of the second part first.

### The DNA of the Expansion

The expression $(2x^2 - \frac{3}{x^2})^6$ follows the general term formula for binomial expansion:
$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

Here, $n=6$, $a=2x^2$, and $b = -\frac{3}{x^2}$. Substituting these values, we obtain:
$$T_{r+1} = \binom{6}{r} (2x^2)^{6-r} \left(-\frac{3}{x^2}ight)^r$$

We isolate the constants and the variables to simplify the expression:
$$T_{r+1} = \binom{6}{r} 2^{6-r} (-3)^r \cdot x^{12-2r} \cdot x^{-2r}$$
$$T_{r+1} = \binom{6}{r} 2^{6-r} (-3)^r \cdot x^{12-4r}$$

### The Fork in the Road

The full expression is $(\frac{1}{60} - \frac{x^8}{81}) \cdot T_{r+1}$. This creates two distinct cases to find the independent term.

**Case 1:** Multiplying $\frac{1}{60}$ by $T_{r+1}$.
For the term to be independent of $x$, the exponent must be zero:
$$12 - 4r = 0 \Rightarrow r = 3$$

Substituting $r=3$ into the coefficient part:
$$	ext{Term}_1 = \frac{1}{60} \cdot \binom{6}{3} \cdot 2^{6-3} \cdot (-3)^3$$
$$	ext{Term}_1 = \frac{1}{60} \cdot 20 \cdot 8 \cdot (-27) = -72$$

### The Final Synthesis

**Case 2:** Multiplying $-\frac{x^8}{81}$ by $T_{r+1}$.
The total power of $x$ becomes $8 + (12 - 4r) = 20 - 4r$. Setting this to zero:
$$20 - 4r = 0 \Rightarrow r = 5$$

Substituting $r=5$ into the coefficient part:
$$	ext{Term}_2 = -\frac{1}{81} \cdot \binom{6}{5} \cdot 2^{6-5} \cdot (-3)^5$$
$$	ext{Term}_2 = -\frac{1}{81} \cdot 6 \cdot 2 \cdot (-243)$$

Since $\frac{243}{81} = 3$, the calculation simplifies:
$$	ext{Term}_2 = -1 \cdot 6 \cdot 2 \cdot (-3) = 36$$

Adding the results from both cases, we find the final value:
$$-72 + 36 = -36$$

The term independent of $x$ is **$-36$**.

Animated Solution for Mathematics - Binomial Theorem: 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 :

Select Answer:

Visualized Solution

Analyze the Expression

Define the General Term $T_{r + 1}$

Substitute Values into $T_{r + 1}$

Simplify the Power of $x$

Distribute the First Bracket

Case 1: Find $r$ for $x^{0}$

Case 1: Calculate Coefficient

Case 2: Find $r$ for $x^{0}$

Case 2: Calculate Coefficient

Final Summation

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The DNA of the Expansion

The Fork in the Road

The Final Synthesis

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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 (601​−81x8​)(2x2−x23​)6 is equal to :

Select Answer:

Visualized Solution

Analyze the Expression

Define the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr+1​

Substitute Values into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr+1​

Simplify the Power of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Distribute the First Bracket

Case 1: Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x0