Find the `3^("rd")` term from the end in the expansion of `(x^(-2//3)-(3)/(x^(2)))^(8).`

Answer

To find the 3rd term from the end in the expansion of \((x^{-\frac{2}{3}} - \frac{3}{x^2})^8\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Terms**:
The total number of terms in the expansion of \((a + b)^n\) is given by \(n + 1\). Here, \(n = 8\), so:
\[ \text{Total terms} = 8 + 1 = 9 \]

2. **Determine the Position of the Required Term**:
The 3rd term from the end is equivalent to the \(9 - 3 + 1 = 7\)th term from the beginning.

3. **Use the Binomial Theorem**:
The general term \(T_r\) in the expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]
Here, \(a = x^{-\frac{2}{3}}\), \(b = -\frac{3}{x^2}\), and \(n = 8\).

4. **Calculate the 7th Term**:
For \(T_7\):
\[ T_7 = \binom{8}{7-1} \left(x^{-\frac{2}{3}}\right)^{8-6} \left(-\frac{3}{x^2}\right)^{6} \]
This simplifies to:
\[ T_7 = \binom{8}{6} \left(x^{-\frac{2}{3}}\right)^{2} \left(-\frac{3}{x^2}\right)^{6} \]

5. **Calculate the Binomial Coefficient**:
\[ \binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]

6. **Simplify the Powers**:
\[ \left(x^{-\frac{2}{3}}\right)^{2} = x^{-\frac{4}{3}} \]
\[ \left(-\frac{3}{x^2}\right)^{6} = (-3)^{6} \cdot \left(\frac{1}{x^2}\right)^{6} = 729 \cdot \frac{1}{x^{12}} = \frac{729}{x^{12}} \]

7. **Combine All Parts**:
Now substituting back:
\[ T_7 = 28 \cdot x^{-\frac{4}{3}} \cdot \frac{729}{x^{12}} = 28 \cdot 729 \cdot x^{-\frac{4}{3} - 12} \]
\[ = 28 \cdot 729 \cdot x^{-\frac{40}{3}} \]

8. **Calculate the Final Coefficient**:
\[ 28 \cdot 729 = 20412 \]

9. **Final Result**:
Thus, the 3rd term from the end in the expansion is:
\[ T_7 = 20412 \cdot x^{-\frac{40}{3}} \]

### Summary:
The 3rd term from the end in the expansion of \((x^{-\frac{2}{3}} - \frac{3}{x^2})^8\) is:
\[ \boxed{20412 \cdot x^{-\frac{40}{3}}} \]