Using binomial theorem, expand each of the following:`(root(3)(x)-root(3)(y))^(6)`

Explanation

To expand the expression \((\sqrt[3]{x} - \sqrt[3]{y})^6\) using the Binomial Theorem, we follow these steps:

### Step 1: Identify the terms in the binomial expansion
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, we have \(a = \sqrt[3]{x}\), \(b = -\sqrt[3]{y}\), and \(n = 6\).

### Step 2: Write the expansion
Using the Binomial Theorem, we can write:
\[ (\sqrt[3]{x} - \sqrt[3]{y})^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt[3]{x})^{6-k} (-\sqrt[3]{y})^k \]

### Step 3: Calculate each term in the expansion
Now we will calculate the terms for \(k = 0\) to \(k = 6\):

1. **For \(k = 0\)**:
\[ \binom{6}{0} (\sqrt[3]{x})^6 (-\sqrt[3]{y})^0 = 1 \cdot x^2 \cdot 1 = x^2 \]

2. **For \(k = 1\)**:
\[ \binom{6}{1} (\sqrt[3]{x})^5 (-\sqrt[3]{y})^1 = 6 \cdot x^{5/3} \cdot (-y^{1/3}) = -6x^{5/3}y^{1/3} \]

3. **For \(k = 2\)**:
\[ \binom{6}{2} (\sqrt[3]{x})^4 (-\sqrt[3]{y})^2 = 15 \cdot x^{4/3} \cdot y^{2/3} = 15x^{4/3}y^{2/3} \]

4. **For \(k = 3\)**:
\[ \binom{6}{3} (\sqrt[3]{x})^3 (-\sqrt[3]{y})^3 = 20 \cdot x \cdot (-y) = -20xy \]

5. **For \(k = 4\)**:
\[ \binom{6}{4} (\sqrt[3]{x})^2 (-\sqrt[3]{y})^4 = 15 \cdot x^{2/3} \cdot y^{4/3} = 15x^{2/3}y^{4/3} \]

6. **For \(k = 5\)**:
\[ \binom{6}{5} (\sqrt[3]{x})^1 (-\sqrt[3]{y})^5 = 6 \cdot x^{1/3} \cdot (-y^{5/3}) = -6x^{1/3}y^{5/3} \]

7. **For \(k = 6\)**:
\[ \binom{6}{6} (\sqrt[3]{x})^0 (-\sqrt[3]{y})^6 = 1 \cdot 1 \cdot y^2 = y^2 \]

### Step 4: Combine all the terms
Now, we combine all the terms we calculated:
\[ (\sqrt[3]{x} - \sqrt[3]{y})^6 = x^2 - 6x^{5/3}y^{1/3} + 15x^{4/3}y^{2/3} - 20xy + 15x^{2/3}y^{4/3} - 6x^{1/3}y^{5/3} + y^2 \]

### Final Answer:
\[ (\sqrt[3]{x} - \sqrt[3]{y})^6 = x^2 - 6x^{5/3}y^{1/3} + 15x^{4/3}y^{2/3} - 20xy + 15x^{2/3}y^{4/3} - 6x^{1/3}y^{5/3} + y^2 \]