1. The problem involves simplifying expressions with cube roots and square roots, such as $$\sqrt[3]{6^{13}} \sqrt{6}$$ and $$\sqrt[3]{6x^2} \cdot \sqrt[3]{25y^3}$$. 2. Recall the properties of radicals: $$\sqrt[n]{a^m} = a^{\frac{m}{n}}$$ and $$\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}$$. 3. Simplify $$\sqrt[3]{6^{13}} \sqrt{6}$$: - Express $$\sqrt{6}$$ as $$6^{\frac{1}{2}}$$. - So the expression becomes $$6^{\frac{13}{3}} \cdot 6^{\frac{1}{2}}$$. - Add exponents: $$6^{\frac{13}{3} + \frac{1}{2}} = 6^{\frac{26}{6} + \frac{3}{6}} = 6^{\frac{29}{6}}$$. 4. Simplify $$\sqrt[3]{6x^2} \cdot \sqrt[3]{25y^3}$$: - Combine under one cube root: $$\sqrt[3]{6x^2 \cdot 25y^3} = \sqrt[3]{150x^2y^3}$$. - Factor inside: $$150 = 25 \times 6$$, but better to look for perfect cubes. - Note $$y^3$$ is a perfect cube, so $$\sqrt[3]{y^3} = y$$. - So expression is $$\sqrt[3]{150x^2} \cdot y$$. 5. Simplify $$\sqrt[3]{54} \cdot \sqrt[3]{25y^2}$$: - Combine: $$\sqrt[3]{54 \cdot 25y^2} = \sqrt[3]{1350y^2}$$. - Factor 1350: $$1350 = 27 \times 50$$, and $$27 = 3^3$$ is a perfect cube. - So $$\sqrt[3]{27 \times 50 y^2} = 3 \sqrt[3]{50 y^2}$$. 6. The other expressions seem incomplete or unclear, so we focus on the above. Final answers: - $$\sqrt[3]{6^{13}} \sqrt{6} = 6^{\frac{29}{6}}$$ - $$\sqrt[3]{6x^2} \cdot \sqrt[3]{25y^3} = y \sqrt[3]{150x^2}$$ - $$\sqrt[3]{54} \cdot \sqrt[3]{25y^2} = 3 \sqrt[3]{50 y^2}$$