Expand `(x^2 -y)^7`

Answer

To expand \((x^2 - y)^7\) using the Binomial Theorem, we will follow these steps:

### Step 1: Identify the components
In the expression \((x^2 - y)^7\), we identify:
- \(a = x^2\)
- \(b = -y\)
- \(n = 7\)

### Step 2: Write the Binomial Expansion Formula
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
Thus, for our case:
\[ (x^2 - y)^7 = \sum_{r=0}^{7} \binom{7}{r} (x^2)^{7-r} (-y)^r \]

### Step 3: Expand the terms
Now we will calculate each term in the expansion for \(r = 0\) to \(r = 7\):

- **For \(r = 0\)**:
\[ \binom{7}{0} (x^2)^{7} (-y)^{0} = 1 \cdot x^{14} \cdot 1 = x^{14} \]

- **For \(r = 1\)**:
\[ \binom{7}{1} (x^2)^{6} (-y)^{1} = 7 \cdot x^{12} \cdot (-y) = -7x^{12}y \]

- **For \(r = 2\)**:
\[ \binom{7}{2} (x^2)^{5} (-y)^{2} = 21 \cdot x^{10} \cdot y^{2} = 21x^{10}y^{2} \]

- **For \(r = 3\)**:
\[ \binom{7}{3} (x^2)^{4} (-y)^{3} = 35 \cdot x^{8} \cdot (-y^{3}) = -35x^{8}y^{3} \]

- **For \(r = 4\)**:
\[ \binom{7}{4} (x^2)^{3} (-y)^{4} = 35 \cdot x^{6} \cdot y^{4} = 35x^{6}y^{4} \]

- **For \(r = 5\)**:
\[ \binom{7}{5} (x^2)^{2} (-y)^{5} = 21 \cdot x^{4} \cdot (-y^{5}) = -21x^{4}y^{5} \]

- **For \(r = 6\)**:
\[ \binom{7}{6} (x^2)^{1} (-y)^{6} = 7 \cdot x^{2} \cdot y^{6} = 7x^{2}y^{6} \]

- **For \(r = 7\)**:
\[ \binom{7}{7} (x^2)^{0} (-y)^{7} = 1 \cdot 1 \cdot (-y^{7}) = -y^{7} \]

### Step 4: Combine all terms
Now we combine all the terms we calculated:
\[ (x^2 - y)^7 = x^{14} - 7x^{12}y + 21x^{10}y^{2} - 35x^{8}y^{3} + 35x^{6}y^{4} - 21x^{4}y^{5} + 7x^{2}y^{6} - y^{7} \]

### Final Answer:
Thus, the expansion of \((x^2 - y)^7\) is:
\[ x^{14} - 7x^{12}y + 21x^{10}y^{2} - 35x^{8}y^{3} + 35x^{6}y^{4} - 21x^{4}y^{5} + 7x^{2}y^{6} - y^{7} \]