Expand :
`(i) (5-3y-2)^(2)` `(ii) (x-(1)/(x)+5)^(2)`

Answer

Let's expand the given expressions step by step.

### (i) Expand \( (5 - 3y - 2)^2 \)

1. **Simplify the expression inside the parentheses:**
\[ 5 - 3y - 2 = 3 - 3y \]
So we rewrite the expression as:
\[ (3 - 3y)^2 \]

2. **Apply the square of a binomial formula:**
The formula \( (a - b)^2 = a^2 - 2ab + b^2 \) applies here where \( a = 3 \) and \( b = 3y \):
\[ (3 - 3y)^2 = 3^2 - 2 \cdot 3 \cdot 3y + (3y)^2 \]

3. **Calculate each term:**
\[ 3^2 = 9 \]
\[ -2 \cdot 3 \cdot 3y = -18y \]
\[ (3y)^2 = 9y^2 \]

4. **Combine the terms:**
\[ 9 - 18y + 9y^2 \]

5. **Rearranging the terms gives:**
\[ 9y^2 - 18y + 9 \]

### Final Result for (i):
\[ (5 - 3y - 2)^2 = 9y^2 - 18y + 9 \]

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### (ii) Expand \( \left( x - \frac{1}{x} + 5 \right)^2 \)

1. **Identify the terms for the binomial formula:**
Let \( a = x - \frac{1}{x} \) and \( b = 5 \).

2. **Apply the square of a binomial formula:**
\[ (a + b)^2 = a^2 + 2ab + b^2 \]
So we have:
\[ \left( x - \frac{1}{x} + 5 \right)^2 = \left( x - \frac{1}{x} \right)^2 + 2 \cdot \left( x - \frac{1}{x} \right) \cdot 5 + 5^2 \]

3. **Calculate \( \left( x - \frac{1}{x} \right)^2 \):**
Using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \):
\[ \left( x - \frac{1}{x} \right)^2 = x^2 - 2 \cdot x \cdot \frac{1}{x} + \left( \frac{1}{x} \right)^2 = x^2 - 2 + \frac{1}{x^2} \]

4. **Calculate \( 2 \cdot \left( x - \frac{1}{x} \right) \cdot 5 \):**
\[ 2 \cdot 5 \cdot \left( x - \frac{1}{x} \right) = 10x - \frac{10}{x} \]

5. **Calculate \( 5^2 \):**
\[ 5^2 = 25 \]

6. **Combine all the terms:**
\[ x^2 - 2 + \frac{1}{x^2} + 10x - \frac{10}{x} + 25 \]

7. **Rearranging gives:**
\[ x^2 + 10x + \left( 25 - 2 \right) + \frac{1}{x^2} - \frac{10}{x} = x^2 + 10x + 23 + \frac{1}{x^2} - \frac{10}{x} \]

### Final Result for (ii):
\[ \left( x - \frac{1}{x} + 5 \right)^2 = x^2 + 10x + 23 - \frac{10}{x} + \frac{1}{x^2} \]

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