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

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 Binomial Expression

First, we need to expand the binomial expression \( \left(2x^2 - \frac{3}{x^2}\right)^6 \) using the Binomial Theorem.

The Binomial Theorem states that:

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

In our case, \( a = 2x^2 \) and \( b = -\frac{3}{x^2} \), and \( n = 6 \):

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

### Step 2: Simplify the Terms

The general term in the expansion will be:

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

This simplifies to:

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

### Step 3: Identify the Independent Term

To find the term independent of \( x \), we set the exponent of \( x \) to zero:

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

### Step 4: Calculate the Coefficient for \( r = 3 \)

Now, we substitute \( r = 3 \) into the expression for \( T_r \):

\[ T_3 = \binom{6}{3} 2^{6-3} (-3)^3 = \binom{6}{3} 2^3 (-27). \]

Calculating \( \binom{6}{3} \):

\[ \binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20. \]

Thus,

\[ T_3 = 20 \cdot 8 \cdot (-27) = -4320. \]

### Step 5: Include the Other Factor

Now we include the factor \( \left(\frac{1}{60} - \frac{x^8}{81}\right) \). The term independent of \( x \) from this factor is \( \frac{1}{60} \).

### Step 6: Combine the Results

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

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

### Step 7: Final Calculation

Now, we also need to consider the term that comes from \( -\frac{x^8}{81} \) multiplied by the term independent of \( x \) from the binomial expansion. The term independent of \( x \) in the expansion of \( \left(2x^2 - \frac{3}{x^2}\right)^6 \) when \( r = 5 \) (because it will give us \( x^0 \) when combined with \( -\frac{x^8}{81} \)) is calculated similarly.

### Conclusion

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

\[ \text{Independent term} = -36. \]