Question

Como resolve isso ??

Answer 1

$\ \ \ (\frac{1}{3})^3+(\frac{1}{4})^2-2^{-3}+16^0\\ \\= \frac{1}{27}+\frac{1}{16}-(\frac{1}{2})^3+1\\ \\= \frac{1}{27}+\frac{1}{16}-\frac{1}{8}+1\\ \\= \frac{16}{432}+\frac{27}{432}-\frac{54}{432}+\frac{432}{432}\\ \\= \frac{16+27-54+432}{432}\\ \\=\frac{421}{432}$