Answer
$17$
Work Step by Step
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:
$=3^2+2^3
\\=9+8
\\=17$
