1. **Problem:** Simplify the radical $\sqrt{63}$. 2. **Formula and rules:** To simplify a square root, factor the number inside into its prime factors and look for perfect squares. Use the property: $$\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$$ If $a$ is a perfect square, then $\sqrt{a} = \text{integer}$. 3. **Factorization:** $$63 = 9 \times 7$$ Since $9$ is a perfect square ($9 = 3^2$), we can write: $$\sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7}$$ 4. **Simplify:** $$\sqrt{9} = 3$$ So, $$\sqrt{63} = 3 \times \sqrt{7} = 3\sqrt{7}$$ 5. **Explanation:** We broke down 63 into 9 and 7 because 9 is a perfect square. Taking the square root of 9 gives 3, which comes out of the radical, leaving $\sqrt{7}$ inside. This is the simplest form because 7 is not a perfect square and cannot be simplified further. **Final answer:** $$\boxed{3\sqrt{7}}$$