1. **State the problem:** Simplify the expression $$\frac{4^3 \div 2^2}{2^0 \times (2^3)^3}$$. 2. **Recall the exponent rules:** - Division of powers with the same base: $$a^m \div a^n = a^{m-n}$$. - Power of a power: $$(a^m)^n = a^{m \times n}$$. - Any number to the zero power is 1: $$a^0 = 1$$. 3. **Simplify the numerator:** $$4^3 \div 2^2 = \frac{4^3}{2^2}$$. Since $4 = 2^2$, rewrite: $$\frac{(2^2)^3}{2^2} = \frac{2^{2 \times 3}}{2^2} = \frac{2^6}{2^2}$$. Using division rule: $$2^{6-2} = 2^4$$. 4. **Simplify the denominator:** $$2^0 \times (2^3)^3 = 1 \times 2^{3 \times 3} = 2^9$$. 5. **Combine numerator and denominator:** $$\frac{2^4}{2^9} = 2^{4-9} = 2^{-5}$$. 6. **Express with positive exponent:** $$2^{-5} = \frac{1}{2^5} = \frac{1}{32}$$. **Final answer:** $$\frac{1}{32}$$.