$$(3^{3}-3^{2}+3-3^{0}):[(2^{4})^{2}:(2^{2})^{3}]$$
$5$
$$\frac{3^{3}-3^{2}+3-3^{0}}{\frac{2^{8}}{\left(2^{2}\right)^{3}}}$$
$$\frac{3^{3}-3^{2}+3-3^{0}}{\frac{2^{8}}{2^{6}}}$$
$$\frac{3^{3}-3^{2}+3-3^{0}}{2^{2}}$$
$$\frac{27-3^{2}+3-3^{0}}{2^{2}}$$
$$\frac{27-9+3-3^{0}}{2^{2}}$$
$$\frac{18+3-3^{0}}{2^{2}}$$
$$\frac{21-3^{0}}{2^{2}}$$
$$\frac{21-1}{2^{2}}$$
$$\frac{20}{2^{2}}$$
$$\frac{20}{4}$$
$$5$$
Show Solution
Hide Solution
