To simplify the term, we are going to make a prime factorization
40 I 2 (We start dividing by 2 since 40 is an even number)
20 I 2 (We continue dividing by 2 since 20 is an even number)
10 I 2 (We continue dividing by 2 since 10 is an even number)
5 I 5 ( We must divide by 5 since 5 is a prime number)
1
Then we rewrite the term as a product of prime numbers and simplify the radical term.
[tex]\begin{gathered} \sqrt[3]{40}=\sqrt[3]{2\cdot2\cdot2\cdot5} \\ \sqrt[3]{2^3\cdot5}\text{ (Writing the product as a power of 2)} \\ \sqrt[3]{2^3}\sqrt[3]{5}\text{ (Separating radicals, using the rule of multiplication)} \\ 2\cdot\sqrt[3]{5}\text{ (Since the index is equal to the power they are cancelled)} \\ \text{The final answer is: } \\ 2\cdot\sqrt[3]{5} \end{gathered}[/tex]
