1. The problem is to simplify the expression $$\frac{(0.5)^3 + (0.05)^3 + (0.005)^3}{(0.1)^3 + (0.01)^3 + (0.001)^3}$$. 2. Recall that to simplify such expressions, we calculate each cube and then sum the results in numerator and denominator separately before dividing. 3. Calculate the cubes: - $$(0.5)^3 = 0.5 \times 0.5 \times 0.5 = 0.125$$ - $$(0.05)^3 = 0.05 \times 0.05 \times 0.05 = 0.000125$$ - $$(0.005)^3 = 0.005 \times 0.005 \times 0.005 = 0.000000125$$ - $$(0.1)^3 = 0.1 \times 0.1 \times 0.1 = 0.001$$ - $$(0.01)^3 = 0.01 \times 0.01 \times 0.01 = 0.000001$$ - $$(0.001)^3 = 0.001 \times 0.001 \times 0.001 = 0.000000001$$ 4. Sum the cubes in numerator: $$0.125 + 0.000125 + 0.000000125 = 0.125125125$$ 5. Sum the cubes in denominator: $$0.001 + 0.000001 + 0.000000001 = 0.001001001$$ 6. Divide numerator sum by denominator sum: $$\frac{0.125125125}{0.001001001} \approx 125$$ 7. Therefore, the simplified value of the expression is $$\boxed{125}$$. This shows how small decimal cubes add up and how division of sums works in such cases.