Question:

Step 1: Find the general term of the binomial expansion $\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$. The general term, $T_{r+1}$, in the expansion of $(a+b)^n$ is given by $T_{r+1} = ^n C_r a^{n-r} b^r$. Here, $a = 2x^2$, $b = -\frac{3}{x^2}$, and $n=6$. $$T_{r+1} = ^{6} C_{r}\left(2 x^{2}\right)^{6-r}\left(-\frac{3}{x^{2}}\right)^{r}$$ Step 2: Simplify the general term to find the power of $x$. $$T_{r+1} = ^{6} C_{r}(2)^{6-r}(x^2)^{6-r}(-3)^{r}(x^{-2})^{r}$$ $$T_{r+1} = ^{6} C_{r}(-3)^{r}(2)^{6-r} x^{2(6-r)} x^{-2r}$$ $$T_{r+1} = ^{6} C_{r}(-3)^{r}(2)^{6-r} x^{12-2r-2r}$$ $$T_{r+1} = ^{6} C_{r}(-3)^{r}(2)^{6-r} x^{12-4r}$$ Step 3: Decompose the given expression to find the term independent of $x$. The given expression is $\left(\frac{1}{60}-\frac{x^{8}}{81}\right) \cdot\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$. To find the term independent of $x$, we consider two parts: 1. The term independent of $x$ in $\frac{1}{60}\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$. 2. The term independent of $x$ in $-\frac{x^{8}}{81}\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$. Step 4: Calculate the term independent of $x$ for the first part: $\frac{1}{60} \cdot T_{r+1}$. For this term to be independent of $x$, the power of $x$ in $T_{r+1}$ must be $0$. So, we set the exponent of $x$ from Step 2 to zero: $12-4r = 0 \implies 4r = 12 \implies r=3$. Substitute $r=3$ into the general term $T_{r+1}$ and multiply by $\frac{1}{60}$: Term$_1 = \frac{1}{60} \cdot ^{6} C_{3}(-3)^{3}(2)^{6-3}$ Term$_1 = \frac{1}{60} \cdot \frac{6 \times 5 \times 4}{3 \times 2 \times 1}(-3)^{3}(2)^{3}$ Term$_1 = \frac{1}{60} \cdot 20 \cdot (-27) \cdot 8$ Term$_1 = \frac{1}{3} \cdot (-27) \cdot 8$ Term$_1 = -9 \cdot 8 = -72$. Step 5: Calculate the term independent of $x$ for the second part: $-\frac{x^{8}}{81} \cdot T_{r+1}$. For this term to be independent of $x$, the total power of $x$ must be $0$. The $x$ term in $-\frac{x^8}{81} \cdot T_{r+1}$ is $x^8 \cdot x^{12-4r}$. So, we set the total exponent of $x$ to zero: $8 + (12-4r) = 0$ $20 - 4r = 0 \implies 4r = 20 \implies r=5$. Substitute $r=5$ into the general term $T_{r+1}$ and multiply by $-\frac{1}{81}$: Term$_2 = -\frac{1}{81} \cdot ^{6} C_{5}(-3)^{5}(2)^{6-5}$ Term$_2 = -\frac{1}{81} \cdot 6 \cdot (-243) \cdot 2$ Note that $243 = 3^5$ and $81 = 3^4$. Term$_2 = -\frac{1}{3^4} \cdot 6 \cdot (-3^5) \cdot 2$ Term$_2 = -1 \cdot 6 \cdot (-3) \cdot 2$ Term$_2 = 18 \cdot 2 = 36$. Step 6: Sum the terms independent of $x$ from both parts. The total term independent of $x$ is the sum of Term$_1$ and Term$_2$: Total term = Term$_1$ + Term$_2$ Total term = $-72 + 36 = -36$. The final answer is $\boxed{-36}$.