1 Simplify without the use of a calculutor- (A) $$ \sqrt[3]{64 x^{6}}-\sqrt[4]{81 x^{8}}+\sqrt{25 x^{4}} $$ (B) $$ \frac{6 \sqrt{2 x^{8}}-2 \sqrt{8 x^{8}}}{\sqrt{8 x^{8}}} $$ (D) $$ \sqrt{x+\sqrt{2 x-1}} \cdot \sqrt{x-\sqrt{2 x-1}} $$

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Let's simplify each expression step-by-step without using a calculator.

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### **A.** $$ \sqrt[3]{64 x^{6}} - \sqrt[4]{81 x^{8}} + \sqrt{25 x^{4}} $$

1. **Simplify each radical individually:**
   - **First Term:** $$ \sqrt[3]{64 x^{6}} $$
     $$
     \sqrt[3]{64} = 4 \quad \text{and} \quad \sqrt[3]{x^6} = x^{2}
     $$
     So, $$ \sqrt[3]{64 x^{6}} = 4x^{2} $$.

- **Second Term:** $$ \sqrt[4]{81 x^{8}} $$
     $$
     \sqrt[4]{81} = 3 \quad \text{and} \quad \sqrt[4]{x^8} = x^{2}
     $$
     So, $$ \sqrt[4]{81 x^{8}} = 3x^{2} $$.

- **Third Term:** $$ \sqrt{25 x^{4}} $$
     $$
     \sqrt{25} = 5 \quad \text{and} \quad \sqrt{x^4} = x^{2}
     $$
     So, $$ \sqrt{25 x^{4}} = 5x^{2} $$.

2. **Combine the simplified terms:**
   $$
   4x^{2} - 3x^{2} + 5x^{2} = (4 - 3 + 5)x^{2} = 6x^{2}
   $$

**Final Simplified Form:**
$$
6x^{2}
$$

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### **B.** $$ \frac{6 \sqrt{2 x^{8}} - 2 \sqrt{8 x^{8}}}{\sqrt{8 x^{8}}} $$

1. **Simplify each radical:**
   - $$ \sqrt{2 x^{8}} = \sqrt{2} \cdot x^{4} $$
   - $$ \sqrt{8 x^{8}} = \sqrt{8} \cdot x^{4} = 2\sqrt{2} \cdot x^{4} $$

2. **Substitute back into the expression:**
   $$
   \frac{6 (\sqrt{2} x^{4}) - 2 (2\sqrt{2} x^{4})}{2\sqrt{2} x^{4}} = \frac{6\sqrt{2} x^{4} - 4\sqrt{2} x^{4}}{2\sqrt{2} x^{4}} = \frac{2\sqrt{2} x^{4}}{2\sqrt{2} x^{4}} = 1
   $$

**Final Simplified Form:**
$$
1
$$

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### **D.** $$ \sqrt{x + \sqrt{2x -1}} \cdot \sqrt{x - \sqrt{2x -1}} $$

1. **Use the property of square roots:**
   $$
   \sqrt{A} \cdot \sqrt{B} = \sqrt{A \cdot B}
   $$
   So,
   $$
   \sqrt{x + \sqrt{2x -1}} \cdot \sqrt{x - \sqrt{2x -1}} = \sqrt{(x + \sqrt{2x -1})(x - \sqrt{2x -1})}
   $$

2. **Simplify the product inside the square root:**
   $$
   (x + \sqrt{2x -1})(x - \sqrt{2x -1}) = x^2 - (2x - 1) = x^2 - 2x + 1 = (x - 1)^2
   $$

3. **Take the square root:**
   $$
   \sqrt{(x - 1)^2} = |x - 1|
   $$
   
   *Note:* The absolute value arises because the square root of a squared term is always non-negative.

**Final Simplified Form:**
$$
|x - 1|
$$

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**Summary of Simplified Expressions:**

- **A.** $$ 6x^{2} $$
- **B.** $$ 1 $$
- **D.** $$ |x - 1| $$
- **A.** $$ 6x^{2} $$ - **B.** $$ 1 $$ - **D.** $$ |x - 1| $$

1 Simplify without the use of a calculutor- (A) \( \sqrt[3]{64 x^{6}}-\sqrt[4]{81 x^{8}}+\sqrt{25 x^{4}} \) (B) \( \frac{6 \sqrt{2 x^{8}}-2 \sqrt{8 x^{8}}}{\sqrt{8 x^{8}}} \) (D) \( \sqrt{x+\sqrt{2 x-1}} \cdot \sqrt{x-\sqrt{2 x-1}} \)

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