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

Answer

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:

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

### Step 2: Expand \(\left(2x^2 - \frac{3}{x^2}\right)^6\)

Using the binomial theorem, we can expand \((a + b)^n\):

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. \]

Here, let \( a = 2x^2 \) and \( b = -\frac{3}{x^2} \):

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

This simplifies to:

\[ = \sum_{r=0}^{6} \binom{6}{r} 2^{6-r} (-3)^r x^{2(6-r) - 2r} = \sum_{r=0}^{6} \binom{6}{r} 2^{6-r} (-3)^r x^{12 - 4r}. \]

### Step 3: Find the term independent of \( x \)

We want the term where the power of \( x \) is zero:

\[ 12 - 4r = 0 \implies 4r = 12 \implies r = 3. \]

### Step 4: Substitute \( r = 3 \) into the expansion

Now, substituting \( r = 3 \):

\[ \text{Term} = \binom{6}{3} 2^{6-3} (-3)^3 = \binom{6}{3} 2^3 (-27). \]

Calculating this:

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

Thus,

\[ \text{Term} = 20 \cdot 8 \cdot (-27) = -4320. \]

### Step 5: Multiply by the constant term \(\frac{1}{60}\)

Now, we multiply this term by the constant term \(\frac{1}{60}\):

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

### Conclusion

The term independent of \( x \) in the expansion is:

\[ \boxed{-72}. \]