1. **Problem statement:** We want to describe how the surface area $O$ of a cube changes when the edge length $a$ is modified. The formula for the surface area is: $$O = 6a^2$$ 2. **Step 1: Doubling the edge length ($a \to 2a$)** Calculate the new surface area: $$O_{new} = 6(2a)^2 = 6 \times 4a^2 = 4 \times 6a^2 = 4O$$ So, the surface area becomes 4 times larger. 3. **Step 2: Quadrupling the edge length ($a \to 4a$)** Calculate the new surface area: $$O_{new} = 6(4a)^2 = 6 \times 16a^2 = 16 \times 6a^2 = 16O$$ The surface area becomes 16 times larger. 4. **Step 3: Halving the edge length ($a \to \frac{a}{2}$)** Calculate the new surface area: $$O_{new} = 6\left(\frac{a}{2}\right)^2 = 6 \times \frac{a^2}{4} = \frac{1}{4} \times 6a^2 = \frac{1}{4}O$$ The surface area becomes one quarter of the original. 5. **Step 4: Increasing the edge length by 10% ($a \to 1.1a$)** Calculate the new surface area: $$O_{new} = 6(1.1a)^2 = 6 \times 1.21a^2 = 1.21 \times 6a^2 = 1.21O$$ The surface area increases by 21%. 6. **Step 5: Multiplying the edge length by $n$ ($a \to na$)** Calculate the new surface area: $$O_{new} = 6(na)^2 = 6n^2a^2 = n^2 \times 6a^2 = n^2O$$ The surface area changes by a factor of $n^2$. **Summary:** - Doubling $a$ quadruples $O$. - Quadrupling $a$ multiplies $O$ by 16. - Halving $a$ divides $O$ by 4. - Increasing $a$ by 10% increases $O$ by 21%. - Multiplying $a$ by $n$ multiplies $O$ by $n^2$.