wing exprertons (without using a calculator). rexpecise 26 - the following surds. (1) \( 25 i^{6} p^{10} \) 2. \( \sqrt[3]{8 x^{12} y^{15}} \) (B) \( 81 \times \sqrt{322^{3} y^{10}} \) 4. \( \sqrt{2 x^{3}} \times \sqrt{8 x^{3}} \) (5) \( (8) \sqrt{2} \cdot \sqrt{8} \) 6. \( \sqrt[6]{64 a^{6} b^{12} c^{18}} \) (0) \( \sqrt[x]{9 x^{10}+16 x^{10}} \) 8. \( \sqrt{\frac{27 x^{3} y^{5}}{12 x^{7} y^{3}}} \) (9) \( \mathrm{g} \sqrt{-} \times \sqrt{27 m} \) (10.) \( \sqrt{2 m^{3}} \times \sqrt{50 m x^{2}} \) (11) \( \frac{\sqrt{4 x^{1}}+\sqrt{4 x^{10}}}{5 x^{3}} \) 12. \( \sqrt[3]{3 x} \times \sqrt[3]{72 x^{5}} \)

Let's simplify each of the given expressions step by step. ### 1. \( 25 i^{6} p^{10} \) First, we simplify \( i^{6} \): - \( i^{2} = -1 \) - \( i^{4} = 1 \) - \( i^{6} = i^{4} \cdot i^{2} = 1 \cdot -1 = -1 \) Now, substituting back: \[ 25 i^{6} p^{10} = 25 \cdot (-1) \cdot p^{10} = -25 p^{10} \] ### 2. \( \sqrt[3]{8 x^{12} y^{15}} \) We can break this down: \[ \sqrt[3]{8} = 2, \quad \sqrt[3]{x^{12}} = x^{4}, \quad \sqrt[3]{y^{15}} = y^{5} \] Thus, \[ \sqrt[3]{8 x^{12} y^{15}} = 2 x^{4} y^{5} \] ### 3. \( 81 \times \sqrt{32^{3} y^{10}} \) First, simplify \( \sqrt{32^{3}} \): \[ 32 = 2^{5} \implies 32^{3} = (2^{5})^{3} = 2^{15} \] So, \[ \sqrt{32^{3}} = \sqrt{2^{15}} = 2^{7.5} = 2^{7} \cdot \sqrt{2} = 128 \sqrt{2} \] Now, substituting back: \[ 81 \times \sqrt{32^{3} y^{10}} = 81 \times 128 \sqrt{2} \times y^{5} = 10368 y^{5} \sqrt{2} \] ### 4. \( \sqrt{2 x^{3}} \times \sqrt{8 x^{3}} \) Using the property of square roots: \[ \sqrt{2 x^{3}} \times \sqrt{8 x^{3}} = \sqrt{(2)(8)(x^{3})(x^{3})} = \sqrt{16 x^{6}} = 4 x^{3} \] ### 5. \( (8) \sqrt{2} \cdot \sqrt{8} \) First, simplify \( \sqrt{8} \): \[ \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2} \] Now substituting back: \[ (8) \sqrt{2} \cdot \sqrt{8} = 8 \sqrt{2} \cdot 2 \sqrt{2} = 16 \cdot 2 = 32 \] ### 6. \( \sqrt[6]{64 a^{6} b^{12} c^{18}} \) Breaking it down: \[ \sqrt[6]{64} = 2, \quad \sqrt[6]{a^{6}} = a, \quad \sqrt[6]{b^{12}} = b^{2}, \quad \sqrt[6]{c^{18}} = c^{3} \] Thus, \[ \sqrt[6]{64 a^{6} b^{12} c^{18}} = 2 a b^{2} c^{3} \] ### 7. \( \sqrt{x}{9 x^{10}+16 x^{10}} \) Combine the terms: \[ 9 x^{10} + 16 x^{10} = 25 x^{10} \] Now, \[ \sqrt{x}{25 x^{10}} = 5 x^{5} \] ### 8. \( \sqrt{\frac{27 x^{3} y^{5}}{12 x^{7} y^{3}}} \) Simplifying the fraction: \[ \frac{27}{12} = \frac{9}{4}, \quad \frac{x^{3}}{x^{7}} = \frac{1}{x^{4}}, \quad \frac{y^{5}}{y^{3}} = y^{2} \] Thus, \[ \sqrt{\frac{27 x^{3} y^{5}}{12 x^{7} y^{3}}} = \sqrt{\frac{9 y^{2}}{4 x^{4}}} = \frac{3 y}{2 x^{2}} \] ### 9. \( g \sqrt{-} \times \sqrt{27 m} \) Assuming \( g \sqrt{-1} = g i \): \[ g i \cdot \sqrt{27 m} = g i \cdot 3 \sqrt{3} \sqrt{m} = 3 g i \sqrt{3 m} \] ### 10. \( \sqrt{2 m^{3}} \times \sqrt{50 m x^{2}} \) Using the property of square roots: \[ \sqrt{2 m^{3}} \times \sqrt{50 m x^{2}} = \sqrt{(2)(50)(m^{3})(m)(x^{2})} = \sqrt{100 m^{4} x^{2}} = 10 m^{2} x \] ### 11. \( \frac{\sqrt{4 x^{1}}+\sqrt{4 x^{10}}}{5 x^{3}} \) Simplifying the square roots: \[ \sqrt{4 x^{1}} = 2 \sqrt{x}, \quad \sqrt{4 x^{10}} = 2 x^{5} \] Thus, \[ \frac{2 \sqrt{x} + 2 x^{5}}{5 x^{3}} = \frac{2(\sqrt{x} + x^{5})}{5 x^{3}} \] ### 12. \( \sqrt[3]{3 x} \times \sqrt[3]{72 x^{5}} \) Using the property of cube roots: \[ \sqrt[3]{3 x \cdot 72 x^{5}} = \sqrt[3]{216 x^{6}} = \sqrt[3]{216} \cdot \sqrt[3]{x^{6}} = 6 x^{2} \] ### Summary of Results 1. \( -25 p^{10} \) 2. \( 2 x^{4} y^{5} \) 3. \( 10368 y^{5} \sqrt{2} \) 4. \( 4 x^{3} \) 5. \( 32 \) 6. \( 2 a b^{2} c^{3} \) 7. \( 5 x^{5} \) 8. \( \frac{3 y}{2 x^{2}} \) 9. \( 3 g i \sqrt{3 m} \) 10. \( 10 m^{2} x \) 11