1. **State the problem:** Simplify the expression $$(4\sqrt{5})^2 + 2^{-3} - 67^0 - \sqrt[3]{\frac{27}{8}}$$. 2. **Recall important rules:** - $(a b)^2 = a^2 b^2$ - $a^{-n} = \frac{1}{a^n}$ - Any number to the zero power is 1, i.e., $a^0 = 1$ - Cube root of a fraction is the cube root of numerator over cube root of denominator 3. **Calculate each term:** - $(4\sqrt{5})^2 = 4^2 \times (\sqrt{5})^2 = 16 \times 5 = 80$ - $2^{-3} = \frac{1}{2^3} = \frac{1}{8}$ - $67^0 = 1$ - $\sqrt[3]{\frac{27}{8}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}} = \frac{3}{2}$ 4. **Substitute back and simplify:** $$80 + \frac{1}{8} - 1 - \frac{3}{2}$$ 5. **Find common denominator 8:** $$80 = \frac{640}{8}, \quad 1 = \frac{8}{8}, \quad \frac{3}{2} = \frac{12}{8}$$ 6. **Rewrite expression:** $$\frac{640}{8} + \frac{1}{8} - \frac{8}{8} - \frac{12}{8} = \frac{640 + 1 - 8 - 12}{8} = \frac{621}{8}$$ 7. **Final answer:** $$\boxed{\frac{621}{8}}$$