Find the
`7^("th")" term of "(1-(x^(2))/(3))^(-4)`

Answer

To find the 7th term of the expansion of \((1 - \frac{x^2}{3})^{-4}\), we can use the Binomial Theorem. The Binomial Theorem states that:

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

For our case, we have \(a = 1\), \(b = -\frac{x^2}{3}\), and \(n = -4\). The general term (the \((r+1)\)th term) in the expansion is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

### Step-by-Step Solution:

1. **Identify the parameters**:
- Here, \(n = -4\), \(a = 1\), and \(b = -\frac{x^2}{3}\).
- We need to find the 7th term, which corresponds to \(r = 6\) (since \(T_{r+1}\) means \(r\) starts from 0).

2. **Write the general term**:
\[ T_{r+1} = \binom{-4}{r} (1)^{-4-r} \left(-\frac{x^2}{3}\right)^r \]

3. **Substituting \(r = 6\)**:
\[ T_{7} = \binom{-4}{6} (1)^{-4-6} \left(-\frac{x^2}{3}\right)^6 \]

4. **Calculate the binomial coefficient**:
The binomial coefficient \(\binom{-n}{r}\) can be calculated using the formula:
\[ \binom{-n}{r} = (-1)^r \binom{n+r-1}{r} \]
So,
\[ \binom{-4}{6} = (-1)^6 \binom{4 + 6 - 1}{6} = \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

5. **Calculate the term**:
\[ T_{7} = 84 \cdot (1)^{-10} \cdot \left(-\frac{x^2}{3}\right)^6 = 84 \cdot \left(-\frac{x^2}{3}\right)^6 \]
\[ = 84 \cdot \left(-1\right)^6 \cdot \frac{x^{12}}{729} = 84 \cdot \frac{x^{12}}{729} \]

6. **Final result**:
\[ T_{7} = \frac{84}{729} x^{12} = \frac{28}{243} x^{12} \]

### Final Answer:
The 7th term of the expansion is \(\frac{28}{243} x^{12}\).