The problem asks us to calculate the cube root of $729$, denoted as $\sqrt[3]{729}$.

Solution:

1. The cube root of a number is the value that, when raised to the power of 3, equals the original number. That is, if $x = \sqrt[3]{729}$, then $x^3 = 729$.

2. Start by factoring $729$: $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$

3. Now, apply the cube root: [ \sqrt[3]{729} = \sqrt[3]{3^6} = 3^{6/3} = 3^2 = 9 ]

Final Answer:

$\boxed{9}$

Would you like a breakdown of any step or further clarification?

Here are 5 related questions you might find interesting:

1. What is the fourth root of 256?
2. How do cube roots relate to exponents and radicals?
3. Can you compute $\sqrt[3]{1728}$?
4. How do you simplify $\sqrt[3]{27x^6}$?
5. What is the cube root of a negative number, such as $\sqrt[3]{-64}$?

Tip: When solving roots or powers, prime factorization can simplify calculations significantly.