## Algebra: A Combined Approach (4th Edition)

$\frac{a^{4}b^{4}}{81y^{4}z^{4}}$
Based on the power of a product rule, we know that $(ab)^{n}=a^{n}b^{n}$ (where $n$ is a positive integer and $a$ and $b$ are real numbers). Therefore, $(\frac{2ab}{6yz})^{4}=\frac{2^{4}a^{4}b^{4}}{6^{4}y^{4}z^{4}}=\frac{16a^{4}b^{4}}{1296y^{4}z^{4}}=\frac{a^{4}b^{4}}{81y^{4}z^{4}}$.