The term independent of `x` in the expansion of `((1)/(60)-(x^(8))/(81)).(2x^(2)-(3)/(x^(2)))^(6)` is equal to:

Options

A. 36
B. -72
C. -36
D. -108

Explanation

To find the term independent of \( x \) in the expansion of

\[ \left(\frac{1}{60} - \frac{x^8}{81}\right) \left(2x^2 - \frac{3}{x^2}\right)^6, \]

we will follow these steps:

### Step 1: Expand the expression
We can rewrite the expression as:

\[ \frac{1}{60} \left(2x^2 - \frac{3}{x^2}\right)^6 - \frac{x^8}{81} \left(2x^2 - \frac{3}{x^2}\right)^6. \]

### Step 2: Use the Binomial Theorem
Using the Binomial Theorem, we can expand \( \left(2x^2 - \frac{3}{x^2}\right)^6 \):

\[ \left(2x^2 - \frac{3}{x^2}\right)^6 = \sum_{r=0}^{6} \binom{6}{r} (2x^2)^r \left(-\frac{3}{x^2}\right)^{6-r}. \]

This simplifies to:

\[ = \sum_{r=0}^{6} \binom{6}{r} 2^r (-3)^{6-r} x^{2r} x^{-2(6-r)} = \sum_{r=0}^{6} \binom{6}{r} 2^r (-3)^{6-r} x^{2(2r - 6)}. \]

### Step 3: Identify the term independent of \( x \)
We need to find the term where the exponent of \( x \) is zero:

\[ 2(2r - 6) = 0 \implies 2r - 6 = 0 \implies r = 3. \]

### Step 4: Calculate the coefficient for \( r = 3 \)
Substituting \( r = 3 \) into the binomial expansion gives:

\[ \binom{6}{3} 2^3 (-3)^{3} = 20 \cdot 8 \cdot (-27) = -4320. \]

### Step 5: Calculate the contribution from both parts
Now, we substitute \( r = 3 \) into both parts of the original expression:

1. From the first part:

\[ \frac{1}{60} \cdot (-4320) = -72. \]

2. From the second part, we need to find the term independent of \( x \) when \( r = 3 \) as well:

The term for \( r = 3 \) in the second part is:

\[ -\frac{x^8}{81} \cdot \binom{6}{3} 2^3 (-3)^{3} = -\frac{x^8}{81} \cdot (-4320). \]

This term will not contribute to the independent term since it involves \( x^8 \).

### Step 6: Combine the contributions
Thus, the only contribution to the independent term comes from the first part:

\[ -72. \]

### Final Answer
The term independent of \( x \) in the expansion is

\[ \boxed{-72}. \]