The problem asks us to reduce the radical $\sqrt{243}$ step by step.

1. Problem Description

The problem asks us to reduce the radical $\sqrt{243}$ step by step.

2. Solution Steps

We can start by finding the prime factorization of

2

4

3. $243 = 3 \times 81 = 3 \times 3 \times 27 = 3 \times 3 \times 3 \times 9 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$.

Therefore, $\sqrt{243} = \sqrt{3^5} = \sqrt{3^4 \times 3} = \sqrt{(3^2)^2 \times 3} = \sqrt{9^2 \times 3}$.

Then, we can simplify the square root as follows:

$\sqrt{243} = \sqrt{81 \times 3} = \sqrt{81} \times \sqrt{3} = 9\sqrt{3}$.

So we have $\sqrt{243} = \sqrt{81} \times \sqrt{3}$. Since $\sqrt{81}=9$, we have $9\sqrt{3}$.

$\sqrt{243} = \sqrt{9 \times 27} = \sqrt{9} \times \sqrt{27} = 3\sqrt{27}$.

Then, $\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}$.

Therefore, $3\sqrt{27} = 3 \times 3\sqrt{3} = 9\sqrt{3}$.

3. Final Answer

$\sqrt{243} = \sqrt{81} \times \sqrt{3} = 9\sqrt{3}$