1. **State the problem:** Compare the value of the 5th root of (2 times the 3rd root of 2) with the 10th root of 6. 2. **Express the roots using exponents:** - The 5th root of a number $x$ is $x^{\frac{1}{5}}$. - The 3rd root of a number $y$ is $y^{\frac{1}{3}}$. 3. **Rewrite the expression:** $$\sqrt[5]{2 \times \sqrt[3]{2}} = \sqrt[5]{2 \times 2^{\frac{1}{3}}} = \sqrt[5]{2^{1 + \frac{1}{3}}} = \sqrt[5]{2^{\frac{4}{3}}}$$ 4. **Simplify the expression inside the 5th root:** $$\sqrt[5]{2^{\frac{4}{3}}} = 2^{\frac{4}{3} \times \frac{1}{5}} = 2^{\frac{4}{15}}$$ 5. **Rewrite the 10th root of 6:** $$\sqrt[10]{6} = 6^{\frac{1}{10}} = (2 \times 3)^{\frac{1}{10}} = 2^{\frac{1}{10}} \times 3^{\frac{1}{10}}$$ 6. **Compare the two expressions:** We want to compare: $$2^{\frac{4}{15}} \quad \text{and} \quad 2^{\frac{1}{10}} \times 3^{\frac{1}{10}}$$ 7. **Divide both sides by $2^{\frac{1}{10}}$ to compare:** $$\frac{2^{\frac{4}{15}}}{2^{\frac{1}{10}}} = 2^{\frac{4}{15} - \frac{1}{10}} = 2^{\frac{8}{30} - \frac{3}{30}} = 2^{\frac{5}{30}} = 2^{\frac{1}{6}}$$ So the comparison reduces to: $$2^{\frac{1}{6}} \quad \text{and} \quad 3^{\frac{1}{10}}$$ 8. **Evaluate approximate values:** - $2^{\frac{1}{6}} \approx 1.122$ - $3^{\frac{1}{10}} \approx 1.116$ Since $1.122 > 1.116$, we have: $$2^{\frac{1}{6}} > 3^{\frac{1}{10}}$$ 9. **Conclusion:** Therefore, $$\sqrt[5]{2 \times \sqrt[3]{2}} > \sqrt[10]{6}$$