We are asked to simplify the fourth root of 243, i.e., $\sqrt[4]{243}$, into its simplified radical form.

Steps:

1. Prime Factorization of 243: $243 \div 3 = 81 \quad \text{(since 243 is divisible by 3)}$ $81 \div 3 = 27$ $27 \div 3 = 9$ $9 \div 3 = 3$ $3 \div 3 = 1$ Therefore, $243 = 3^5$.

2. Applying the Fourth Root: We are taking the fourth root of $3^5$: [ \sqrt[4]{243} = \sqrt[4]{3^5} ] We can separate this into: [ \sqrt[4]{3^4} \times \sqrt[4]{3} ] Since $\sqrt[4]{3^4} = 3$, the expression simplifies to: [ 3 \times \sqrt[4]{3} = 3 \sqrt[4]{3} ]

Thus, the simplified radical form of $\sqrt[4]{243}$ is: $3 \sqrt[4]{3}$

Would you like further clarification or details on any step?

Here are some related questions:

1. How do you simplify cube roots?
2. What is the prime factorization method?
3. How do you handle higher-order roots in different types of problems?
4. Can you explain the general process for simplifying any radical?
5. What happens when the index of the root and the exponent are not perfectly divisible?

Tip: When simplifying radicals, always start with prime factorization for better clarity.