1. **State the problem:** Simplify and verify the equality: $$\sqrt{49} \cdot \sqrt[3]{14} \cdot \sqrt[5]{48} = \sqrt[10]{6^7} \cdot \sqrt[6]{7^5} \cdot \sqrt[15]{2^{14}}$$ 2. **Recall the root and exponent rules:** - The $n$th root of $a$ is $a^{\frac{1}{n}}$. - Multiplying roots corresponds to multiplying their expressions with fractional exponents. - To compare both sides, express all terms with fractional exponents and simplify. 3. **Convert left side to fractional exponents:** - $\sqrt{49} = 49^{\frac{1}{2}}$ - $\sqrt[3]{14} = 14^{\frac{1}{3}}$ - $\sqrt[5]{48} = 48^{\frac{1}{5}}$ 4. **Simplify $49^{\frac{1}{2}}$:** - $49 = 7^2$, so $$49^{\frac{1}{2}} = (7^2)^{\frac{1}{2}} = 7^{2 \cdot \frac{1}{2}} = 7^1 = 7$$ 5. **Prime factorize 14 and 48:** - $14 = 2 \cdot 7$ - $48 = 2^4 \cdot 3$ 6. **Rewrite left side with prime factors and fractional exponents:** $$7 \cdot (2 \cdot 7)^{\frac{1}{3}} \cdot (2^4 \cdot 3)^{\frac{1}{5}} = 7 \cdot 2^{\frac{1}{3}} \cdot 7^{\frac{1}{3}} \cdot 2^{\frac{4}{5}} \cdot 3^{\frac{1}{5}}$$ 7. **Combine like bases on the left side:** - For base 7: $$7^1 \cdot 7^{\frac{1}{3}} = 7^{1 + \frac{1}{3}} = 7^{\frac{4}{3}}$$ - For base 2: $$2^{\frac{1}{3}} \cdot 2^{\frac{4}{5}} = 2^{\frac{1}{3} + \frac{4}{5}} = 2^{\frac{5}{15} + \frac{12}{15}} = 2^{\frac{17}{15}}$$ - For base 3: $$3^{\frac{1}{5}}$$ So left side is: $$7^{\frac{4}{3}} \cdot 2^{\frac{17}{15}} \cdot 3^{\frac{1}{5}}$$ 8. **Convert right side to fractional exponents:** - $\sqrt[10]{6^7} = 6^{\frac{7}{10}}$ - $\sqrt[6]{7^5} = 7^{\frac{5}{6}}$ - $\sqrt[15]{2^{14}} = 2^{\frac{14}{15}}$ 9. **Prime factorize 6:** - $6 = 2 \cdot 3$ 10. **Rewrite right side with prime factors:** $$6^{\frac{7}{10}} \cdot 7^{\frac{5}{6}} \cdot 2^{\frac{14}{15}} = (2 \cdot 3)^{\frac{7}{10}} \cdot 7^{\frac{5}{6}} \cdot 2^{\frac{14}{15}} = 2^{\frac{7}{10}} \cdot 3^{\frac{7}{10}} \cdot 7^{\frac{5}{6}} \cdot 2^{\frac{14}{15}}$$ 11. **Combine like bases on the right side:** - For base 2: $$2^{\frac{7}{10}} \cdot 2^{\frac{14}{15}} = 2^{\frac{7}{10} + \frac{14}{15}} = 2^{\frac{21}{30} + \frac{28}{30}} = 2^{\frac{49}{30}}$$ - For base 3: $$3^{\frac{7}{10}}$$ - For base 7: $$7^{\frac{5}{6}}$$ So right side is: $$2^{\frac{49}{30}} \cdot 3^{\frac{7}{10}} \cdot 7^{\frac{5}{6}}$$ 12. **Compare exponents of each base on both sides:** - Base 7: Left: $\frac{4}{3} = \frac{8}{6}$ Right: $\frac{5}{6}$ - Base 2: Left: $\frac{17}{15} = \frac{34}{30}$ Right: $\frac{49}{30}$ - Base 3: Left: $\frac{1}{5} = \frac{6}{30}$ Right: $\frac{7}{10} = \frac{21}{30}$ 13. **Conclusion:** The exponents for bases 7, 2, and 3 are not equal on both sides, so the equality does not hold as stated. **Final answer:** $$\sqrt{49} \cdot \sqrt[3]{14} \cdot \sqrt[5]{48} \neq \sqrt[10]{6^7} \cdot \sqrt[6]{7^5} \cdot \sqrt[15]{2^{14}}$$