Key idea. Use the sum-of-cubes identity with reciprocal terms.

1. For any u and v, expanding \((u+v)^3\) gives

   \[u^3+v^3=(u+v)^3-3uv(u+v).\]

2. Here \(u=x\), \(v=1/x\). We know \(u+v=3\) and \(uv=1\).

3. Substitution gives

   \[x^3+\frac1{x^3}=3^3-3\cdot1\cdot3=27-9=18.\]

Remember. Knowing the sum and product of two quantities is enough to find their sum of cubes.