1. The problem is to simplify the expression $\sqrt[3]{18}$, which means finding the cube root of 18. 2. Recall the cube root property: $\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$. We want to factor 18 into a product where one factor is a perfect cube. 3. Factor 18: $18 = 9 \times 2 = 3^2 \times 2$. 4. Since $3^2$ is not a perfect cube, check if we can rewrite 18 as $3^3 \times \frac{18}{3^3}$, but $3^3 = 27$ which is greater than 18, so no. 5. Instead, rewrite 18 as $\sqrt[3]{9 \times 2} = \sqrt[3]{9} \times \sqrt[3]{2}$. 6. Note that 9 is $3^2$, not a perfect cube, so no further simplification there. 7. However, we can try to express 18 as $\sqrt[3]{(3^3) \times \frac{18}{27}}$ but $\frac{18}{27} = \frac{2}{3}$ which is not an integer. 8. Since no perfect cube factors exist in 18, the simplest radical form is $\sqrt[3]{18}$ itself. 9. Alternatively, approximate or leave as is. But if we want to express in simplest radical form, factor out cube roots of perfect cubes only. Final answer: $\sqrt[3]{18}$