Write the general term in the expansion of `(x^(2)-y)^(6)`.

Explanation

To find the general term in the expansion of \((x^2 - y)^6\), we will use the Binomial Theorem. The Binomial Theorem states that:

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

where \(\binom{n}{r}\) is the binomial coefficient, \(a\) and \(b\) are the terms in the binomial expression, and \(n\) is the exponent.

### Step-by-Step Solution:

1. **Identify the terms**: In our case, we have:
- \(a = x^2\)
- \(b = -y\)
- \(n = 6\)

2. **Write the general term**: The general term \(T_{r+1}\) in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

3. **Substitute the values**: Substituting \(a\), \(b\), and \(n\) into the formula:
\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-y)^r \]

4. **Simplify the expression**:
- The term \((x^2)^{6-r} = x^{2(6-r)} = x^{12 - 2r}\)
- The term \((-y)^r = (-1)^r y^r\)

Therefore, we can rewrite the general term as:
\[ T_{r+1} = \binom{6}{r} x^{12 - 2r} (-1)^r y^r \]

5. **Final expression**: Combining all parts, the general term \(T_{r+1}\) is:
\[ T_{r+1} = (-1)^r \binom{6}{r} x^{12 - 2r} y^r \]

### Final Answer:
The general term in the expansion of \((x^2 - y)^6\) is:
\[ T_{r+1} = (-1)^r \binom{6}{r} x^{12 - 2r} y^r \]