To simplify $\sqrt[3]{40}$, we begin by finding the prime factorization of 40.

$40 = 2^3 \times 5$

Now, apply the cube root:

[ \sqrt[3]{40} = \sqrt[3]{2^3 \times 5} = \sqrt[3]{2^3} \times \sqrt[3]{5} ]

The cube root of $2^3$ simplifies to 2:

[ \sqrt[3]{40} = 2 \times \sqrt[3]{5} = 2\sqrt[3]{5} ]

Thus, the simplified form of $\sqrt[3]{40}$ is:

$2\sqrt[3]{5}$

Would you like further details on cube roots or any other related concepts?

Here are 5 relative questions to expand your understanding:

1. How do you simplify expressions with higher-order roots, like the 4th root?
2. What is the difference between cube roots and square roots in simplification?
3. How would you handle the cube root of a fraction like $\sqrt[3]{\frac{27}{8}}$?
4. Can you approximate cube roots of non-perfect cubes without simplification?
5. How are cube roots used in solving cubic equations?

Tip: Always look for perfect powers in a number's prime factorization when simplifying roots.