- Break the expression into its individual components: \(\sqrt[3]{64} \cdot \sqrt[3]{a^6} \cdot \sqrt[3]{b^7} \cdot \sqrt[3]{c^9}\).
- Calculate the cube root of the constant and simplify each variable term by dividing exponents by the root index: \(4 \cdot a^2 \cdot (b^2 \sqrt[3]{b}) \cdot c^3\).
- Combine the results to get the simplified expression: \(4 a^2 b^2 c^3 \sqrt[3]{b}\).
- Match the result to the correct option: \(\boxed{4 a^2 b^2 c^3 (\sqrt[3]{b})}\).
Explanation
- Breaking down the Radicand
To simplify the expression \(\sqrt[3]{64 a^6 b^7 c^9}\), we use the property that \(\sqrt[n]{x \cdot y \cdot z} = \sqrt[n]{x} \cdot \sqrt[n]{y} \cdot \sqrt[n]{z}\) and \(\sqrt[n]{x^m} = x^{m/n}\). We break the radicand into its individual factors: \[\sqrt[3]{64 a^6 b^7 c^9} = \sqrt[3]{64} \cdot \sqrt[3]{a^6} \cdot \sqrt[3]{b^7} \cdot \sqrt[3]{c^9}.\]
- Simplifying Each Factor
Now, evaluate each part separately:
- For the constant: \(\sqrt[3]{64} = 4\).
- For the variable \(a\): \(\sqrt[3]{a^6} = a^{6/3} = a^2\).
- For the variable \(b\): We express \(b^7\) as \(b^6 \cdot b^1\), so \(\sqrt[3]{b^7} = \sqrt[3]{b^6} \cdot \sqrt[3]{b} = b^{6/3} \cdot \sqrt[3]{b} = b^2 \sqrt[3]{b}\).
- For the variable \(c\): \(\sqrt[3]{c^9} = c^{9/3} = c^3\).
- Combining and Comparing with Options
Combine all the simplified parts:
\[4 \cdot a^2 \cdot (b^2 \sqrt[3]{b}) \cdot c^3 = 4 a^2 b^2 c^3 \sqrt[3]{b}.\]
Comparing this to the given options, we find it exactly matches the second option.
- Final Result
The expression simplifies to \(4 a^2 b^2 c^3 \sqrt[3]{b}\). The final answer is \(\boxed{4 a^2 b^2 c^3 (\sqrt[3]{b})}\).
Examples
Simplifying radical expressions is a foundational skill in algebra, used in various fields like physics and engineering. For example, when calculating the volume of an object or determining the dimensions of a shape from its volume (e.g., finding the side length of a cube from its volume), you often need to simplify expressions involving roots. This allows you to work with cleaner, more manageable numbers in your formulas, ensuring accuracy in physical calculations.