Select the simplification that accurately explains the following statement. \( \sqrt[8]{7}=7^{\frac{1}{3}} \) A. \( \left(7^{\frac{1}{3}}\right)^{8}=7^{\frac{1}{3}} \cdot 7^{\frac{1}{3}} \cdot 7^{\frac{1}{3}}=7 \cdot\left(\frac{1}{3}+\frac{1}{8}+\frac{1}{3}\right)=7 \cdot \frac{3}{3}=7 \cdot 1=7 \) B. \( \left(7^{\frac{1}{3}}\right)^{3}=7^{\frac{1}{3}} \cdot 7^{\frac{1}{3}} \cdot 7^{\frac{1}{3}}=7 \cdot 7^{\frac{1}{3}}=3 \cdot \frac{1}{3} \cdot 7=1 \cdot 7=7 \) C. \( \left(7^{\frac{1}{3}}\right)^{8}=7^{\frac{1}{3}} \cdot 7^{\frac{1}{3}} \cdot 7^{\frac{1}{3}}=7^{\frac{1}{3} \cdot \frac{1}{3}} \frac{1}{3}=7^{\frac{3}{3}}=7^{1}=7 \) D. \( \left(7^{\frac{1}{3}}\right)^{8}=7^{\frac{1}{3}} \cdot 7^{\frac{1}{3}} \cdot 7^{\frac{1}{3}}=7^{\frac{1}{3}}+\frac{1}{3}+\frac{1}{3}=7^{\frac{3}{3}}=7^{1}=7 \)
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Bonus Knowledge
Did you know that the roots and exponents we use today have roots (pun intended!) deeply embedded in history? The concept of roots can be traced back to ancient civilizations. The Babylonians had methods for approximating square roots, and by the time of the Greeks, mathematicians like Euclid were rigorously defining concepts of ratios that related to roots, laying groundwork for our current understanding. Talk about a journey through time! When dealing with exponents, a common mistake is forgetting the power of a power rule, which states that when you raise a power to another power, you multiply the exponents. For instance, if you see \( \left(7^{\frac{1}{3}}\right)^8 \), don't forget to multiply \(\frac{1}{3}\) by \(8\), which gives you \(7^{\frac{8}{3}}\), not some addition mishap or an unfavorable mix-up with fractions! Keep it straightforward, and the math will be your ally!