## Intermediate Algebra for College Students (7th Edition)

$17$
RECALL: $a^{\frac{m}{n}}=\sqrt[n]{a^m}$ For the first term, $a=27$, $m=2$ and $n=3$. For the second term, $a=16$, $m=3$ and $n=4$. Use the rule above to have: $=\sqrt[3]{27^2}+\sqrt[4]{16^3}$ Write 27 as a power of 3 and 16 as a power of 2 to obtain: $=\sqrt[3]{(3^3)^2}+\sqrt[4]{(2^4)^3}$ Use the rule $(a^m)^n=a^{mn}$ to obtain: $=\sqrt[3]{3^{3(2)}}+\sqrt[4]{2^{4(3)}} \\=\sqrt[3]{3^{6}}+ \sqrt[4]{2^{12}} \\=\sqrt[3]{(3^2)^3}+\sqrt[4]{(2^3)^4}$ Use the rule $\sqrt[n]{a^n} = a, a\ge 0$ to obtain: $=3^2+2^3 \\=9+8 \\=17$