1. Key Concept: The Binomial Theorem and Finding the Term Independent of $x$

The Binomial Theorem allows us to expand expressions of the form $(a+b)^n$. The general term, often denoted as $T_{r+1}$, in the expansion of $(a+b)^n$ is given by: $T_{r+1} = {}^nC_r \, a^{n-r} \, b^r$ where $r$ is an integer ranging from $0$ to $n$.

To find the term independent of $x$ (also known as the constant term), our goal is to identify the specific value of $r$ for which the exponent of $x$ in the general term becomes zero. Once this value of $r$ is determined, we substitute it back into the general term formula to calculate the numerical value of the constant term.

2. Analyzing the Given Expression and Strategic Breakdown

The given expression is a product of two factors: $\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$ To find the term independent of $x$ in this product, we first distribute the terms from the first factor into the binomial expansion of the second factor. This approach simplifies the problem by breaking it down into two more manageable parts:

• Part 1: Finding the term independent of $x$ in ${1 \over {60}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$
• Part 2: Finding the term independent of $x$ in $- {{{x^8}} \over {81}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$

We will calculate the term independent of $x$ for each part separately and then sum these results to get the final answer.

3. Determining the General Term for the Binomial Expansion ${\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$

Let's first find the general term, $T_{r+1}$, for the common binomial expansion ${\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$. Here, we have:

• $a = 2x^2$
• $b = -{3 \over {{x^2}}} = -3x^{-2}$
• $n = 6$

Applying the general term formula $T_{r+1} = {}^nC_r \, a^{n-r} \, b^r$: $T_{r+1} = {}^6C_r \, (2x^2)^{6-r} \, (-3x^{-2})^r$ Now, we meticulously separate the constant coefficients and the terms involving $x$ to simplify: $T_{r+1} = {}^6C_r \, (2)^{6-r} \, (x^2)^{6-r} \, (-3)^r \, (x^{-2})^r$ $T_{r+1} = {}^6C_r \, 2^{6-r} \, (-3)^r \, x^{2(6-r)} \, x^{-2r}$ To find the net power of $x$, we combine the exponents: $T_{r+1} = {}^6C_r \, 2^{6-r} \, (-3)^r \, x^{12-2r-2r}$ $T_{r+1} = {}^6C_r \, 2^{6-r} \, (-3)^r \, x^{12-4r}$ This expression represents the general term for the expansion of ${\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$.

4. Finding the Term Independent of $x$ in Part 1: ${1 \over {60}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$

For a term to be independent of $x$, its exponent must be $0$. From the general term derived above, the power of $x$ is $12-4r$. So, we set the exponent to zero and solve for $r$: $12-4r = 0$ $4r = 12$ $r = 3$ Since $r=3$ is a valid integer between $0$ and $6$, this term exists in the expansion.

Now, we substitute $r=3$ into the general term (omitting $x^0$) and multiply by the constant factor ${1 \over {60}}$: Term independent of $x$ in Part 1 = ${1 \over {60}} \times \left( {}^6C_3 \, 2^{6-3} \, (-3)^3 \right)$ $= {1 \over {60}} \times \left( {}^6C_3 \, 2^3 \, (-3)^3 \right)$ Let's calculate the individual components:

• $^6C_3 = {6 \times 5 \times 4 \over 3 \times 2 \times 1} = 20$
• $2^3 = 8$
• $(-3)^3 = -27$ Substitute these values back: $= {1 \over {60}} \times (20 \times 8 \times (-27))$ $= {1 \over {60}} \times (160 \times (-27))$ $= {1 \over {60}} \times (-4320)$ $= -72$

5. Finding the Term Independent of $x$ in Part 2: $- {{{x^8}} \over {81}}{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$

In this part, the overall term includes an additional factor of $x^8$ from the initial distribution: $- {1 \over {81}} \times x^8 \times \left( {}^6C_r \, 2^{6-r} \, (-3)^r \, x^{12-4r} \right)$ The combined power of $x$ in this expression is $x^8 \cdot x^{12-4r} = x^{8+12-4r} = x^{20-4r}$. To find the term independent of $x$, we set this combined exponent to $0$: $20-4r = 0$ $4r = 20$ $r = 5$ Again, $r=5$ is a valid integer between $0$ and $6$, so this term exists.

Now, we substitute $r=5$ into the general term (excluding the $x$ part, which becomes $x^0$) and multiply by the constant factor $- {1 \over {81}}$: Term independent of $x$ in Part 2 = $- {1 \over {81}} \times \left( {}^6C_5 \, 2^{6-5} \, (-3)^5 \right)$ $= - {1 \over {81}} \times \left( {}^6C_5 \, 2^1 \, (-3)^5 \right)$ Let's calculate the individual components:

• $^6C_5 = {}^6C_{6-5} = {}^6C_1 = 6$
• $2^1 = 2$
• $(-3)^5 = -243$ Substitute these values back: $= - {1 \over {81}} \times (6 \times 2 \times (-243))$ $= - {1 \over {81}} \times (12 \times (-243))$ $= - {1 \over {81}} \times (-2916)$ $= 36$ Tip for JEE Aspirants: Be extremely careful with signs! A double negative, as seen here ($- \frac{1}{81} \times (-2916)$), correctly results in a positive value. Mistakes with signs are common in such calculations.

6. Summing the Independent Terms

The total term independent of $x$ in the original expression is the sum of the independent terms from Part 1 and Part 2: Total term = (Term from Part 1) + (Term from Part 2) Total term = $-72 + 36$ Total term = $-36$

7. Tips for JEE Aspirants

• Break Down Complex Problems: Always aim to simplify complex expressions by breaking them into smaller, more manageable parts. This strategy, especially distributing products, significantly reduces the chances of errors.
• Systematic Use of General Term: The general term formula ($T_{r+1}$) is a powerful tool. Use it systematically to handle powers of variables and coefficients. Avoid trying to guess the term directly.
• Exponent Rules are Crucial: A solid understanding of exponent rules (e.g., $x^a \cdot x^b = x^{a+b}$ and $(x^a)^b = x^{ab}$) is fundamental. Minor errors here can lead to incorrect values of $r$.
• Careful with Combinations ($^nC_r$) and Powers: Ensure accurate calculation of binomial coefficients and powers, especially negative bases raised to odd or even powers.

8. Summary and Key Takeaway

This problem effectively tests the application of the Binomial Theorem for finding a term independent of $x$ in a more complex product expression. The key to solving such problems lies in a systematic approach: first, distribute to separate the terms; second, apply the general term formula to each part; third, set the power of $x$ to zero to find the appropriate $r$; and finally, sum the constant terms. Careful algebraic manipulation and attention to signs are paramount for arriving at the correct solution.