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

Explanation

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

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

For our case, we can identify \(p = x^2\), \(q = -y\), and \(n = 6\).

### Step 1: Identify the general term
The general term in the expansion can be expressed as:

\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]

Substituting our values:

\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-y)^r \]

### Step 2: Simplify the general term
Now, we simplify the expression:

\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-y)^r = \binom{6}{r} x^{2(6-r)} (-y)^r \]

This simplifies to:

\[ T_{r+1} = \binom{6}{r} x^{12 - 2r} (-1)^r y^r \]

### Step 3: Final expression of the general term
Thus, the general term \(T_{r+1}\) can be written as:

\[ T_{r+1} = \binom{6}{r} (-1)^r x^{12 - 2r} y^r \]

### Conclusion
The general term in the expansion of \((x^2 - y)^6\) is:

\[ T_{r+1} = \binom{6}{r} (-1)^r x^{12 - 2r} y^r \]