- Distribute the exponent \(\frac{1}{5}\) to all terms: \((32^2)^{\frac{1}{5}} (x^4)^{\frac{1}{5}} (y^{12})^{\frac{1}{5}}\).
- Simplify the coefficient: \((32^2)^{\frac{1}{5}} = (2^{10})^{\frac{1}{5}} = 4\).
- Simplify the variables: \(x^{\frac{4}{5}} \cdot y^{\frac{12}{5}} = x^{\frac{4}{5}} \cdot y^2 \cdot y^{\frac{2}{5}}\).
- Combine into the final form: \(\boxed{4 y^2 \sqrt[5]{x^4 y^2}}\).
Explanation
- Distributing the Exponent
The goal is to simplify the expression \((32^2 x^4 y^{12})^{\frac{1}{5}}\). We apply the exponent laws for powers of products: \((abc)^n = a^n b^n c^n\) and \((a^m)^n = a^{mn}\). Applying these to our expression, we distribute the exponent \(\frac{1}{5}\) to each term inside the parentheses:
\[(32^2)^{\frac{1}{5}} \cdot (x^4)^{\frac{1}{5}} \cdot (y^{12})^{\frac{1}{5}}\]
- Simplifying Each Term
Next, we simplify each of the three parts:
- For the numerical coefficient: \(32 = 2^5\), so \(32^2 = (2^5)^2 = 2^{10}\). Thus, \((32^2)^{\frac{1}{5}} = (2^{10})^{\frac{1}{5}} = 2^{10 \cdot \frac{1}{5}} = 2^2 = 4\).
- For the variable \(x\): \((x^4)^{\frac{1}{5}} = x^{4 \cdot \frac{1}{5}} = x^{\frac{4}{5}}\).
- For the variable \(y\): \((y^{12})^{\frac{1}{5}} = y^{12 \cdot \frac{1}{5}} = y^{\frac{12}{5}}\). Since \(\frac{12}{5} = 2 + \frac{2}{5}\), we can rewrite this as \(y^2 \cdot y^{\frac{2}{5}}\).
- Combining and Formatting the Result
Now, combine all the simplified parts:
\[4 \cdot x^{\frac{4}{5}} \cdot y^2 \cdot y^{\frac{2}{5}}\]
Rearrange the terms for better readability:
\[4 y^2 \cdot x^{\frac{4}{5}} \cdot y^{\frac{2}{5}}\]
Finally, combine the terms with fractional exponents into a single fifth root:
\[x^{\frac{4}{5}} \cdot y^{\frac{2}{5}} = (x^4 y^2)^{\frac{1}{5}} = \sqrt[5]{x^4 y^2}\]
Thus, the fully simplified expression is \(4 y^2 \sqrt[5]{x^4 y^2}\). This matches the second option provided.
Examples
Simplifying expressions with powers and roots is a fundamental skill in algebra, essential for solving complex equations in fields like physics or engineering. For example, if you are modeling the volume of an object where dimensions are given in high-order powers, converting back to a standard root form allows you to more easily compare the dimensions or interpret the physical values. This technique ensures calculations remain manageable and accurate.