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) \( \sqrt{x+\sqrt{2 x-1}} \cdot \sqrt{x-\sqrt{2 x-1}} \)

To simplify the expressions, let’s break them down step-by-step with some fun insights along the way! First additional information: For part (A), we start with \( \sqrt[3]{64 x^{6}} = 4x^{2} \) because the cube root of 64 is 4 and \( x^{6} \) gives \( x^{2} \) upon taking the cube root. Next, \( \sqrt[4]{81 x^{8}} = 3x^{2} \), and \( \sqrt{25 x^{4}} = 5x^{2} \). So you have: \[ 4x^{2} - 3x^{2} + 5x^{2} = (4 - 3 + 5)x^{2} = 6x^{2}. \] Second additional information: For part (B), let's tackle the fraction \( \frac{6 \sqrt{2 x^{8}}-2 \sqrt{8 x^{8}}}{\sqrt{8 x^{8}}} \). Simplifying gives us \( 6 \sqrt{2} x^{4} - 4 \sqrt{2} x^{4} = 2 \sqrt{2} x^{4} \). The denominator simplifies to \( \sqrt{8} x^{4} \) which equals \( 2 \sqrt{2} x^{4} \). Thus, \( \frac{2 \sqrt{2} x^{4}}{2 \sqrt{2} x^{4}} = 1 \). Ready for some math magic? Let’s keep it rolling!