Expand using binomial theorem:
`(i) (1-2x)^(4) " " (ii) (1+(1)/(x^(2)))^(4)`

Explanation

To solve the given problems using the Binomial Theorem, we will apply the theorem step by step for each part.

### Part (i): Expand \( (1 - 2x)^4 \)

**Step 1: Identify the components**
- Here, \( a = 1 \), \( b = -2x \), and \( n = 4 \).

**Step 2: Write the Binomial Expansion Formula**
The Binomial Theorem states:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our case:
\[ (1 - 2x)^4 = \sum_{k=0}^{4} \binom{4}{k} (1)^{4-k} (-2x)^k \]

**Step 3: Calculate each term of the expansion**
- For \( k = 0 \):
\[ \binom{4}{0} (1)^{4} (-2x)^{0} = 1 \]
- For \( k = 1 \):
\[ \binom{4}{1} (1)^{3} (-2x)^{1} = 4 \cdot (-2x) = -8x \]
- For \( k = 2 \):
\[ \binom{4}{2} (1)^{2} (-2x)^{2} = 6 \cdot (4x^2) = 24x^2 \]
- For \( k = 3 \):
\[ \binom{4}{3} (1)^{1} (-2x)^{3} = 4 \cdot (-8x^3) = -32x^3 \]
- For \( k = 4 \):
\[ \binom{4}{4} (1)^{0} (-2x)^{4} = 1 \cdot 16x^4 = 16x^4 \]

**Step 4: Combine all terms**
Putting it all together:
\[ (1 - 2x)^4 = 1 - 8x + 24x^2 - 32x^3 + 16x^4 \]

### Part (ii): Expand \( \left(1 + \frac{1}{x^2}\right)^4 \)

**Step 1: Identify the components**
- Here, \( a = 1 \), \( b = \frac{1}{x^2} \), and \( n = 4 \).

**Step 2: Write the Binomial Expansion Formula**
Using the same Binomial Theorem:
\[ \left(1 + \frac{1}{x^2}\right)^4 = \sum_{k=0}^{4} \binom{4}{k} (1)^{4-k} \left(\frac{1}{x^2}\right)^k \]

**Step 3: Calculate each term of the expansion**
- For \( k = 0 \):
\[ \binom{4}{0} (1)^{4} \left(\frac{1}{x^2}\right)^{0} = 1 \]
- For \( k = 1 \):
\[ \binom{4}{1} (1)^{3} \left(\frac{1}{x^2}\right)^{1} = 4 \cdot \frac{1}{x^2} = \frac{4}{x^2} \]
- For \( k = 2 \):
\[ \binom{4}{2} (1)^{2} \left(\frac{1}{x^2}\right)^{2} = 6 \cdot \frac{1}{x^4} = \frac{6}{x^4} \]
- For \( k = 3 \):
\[ \binom{4}{3} (1)^{1} \left(\frac{1}{x^2}\right)^{3} = 4 \cdot \frac{1}{x^6} = \frac{4}{x^6} \]
- For \( k = 4 \):
\[ \binom{4}{4} (1)^{0} \left(\frac{1}{x^2}\right)^{4} = 1 \cdot \frac{1}{x^8} = \frac{1}{x^8} \]

**Step 4: Combine all terms**
Putting it all together:
\[ \left(1 + \frac{1}{x^2}\right)^4 = 1 + \frac{4}{x^2} + \frac{6}{x^4} + \frac{4}{x^6} + \frac{1}{x^8} \]

### Final Answers:
1. \( (1 - 2x)^4 = 1 - 8x + 24x^2 - 32x^3 + 16x^4 \)
2. \( \left(1 + \frac{1}{x^2}\right)^4 = 1 + \frac{4}{x^2} + \frac{6}{x^4} + \frac{4}{x^6} + \frac{1}{x^8} \)