The equation you're trying to solve is:

$0 = \sqrt{64 - x^3}$

Step 1: Square both sides of the equation to eliminate the square root.

Since $\sqrt{64 - x^3} = 0$, squaring both sides gives:

$0^2 = (\sqrt{64 - x^3})^2$ $0 = 64 - x^3$

Step 2: Solve for $x^3$.

Now, isolate $x^3$:

$x^3 = 64$

Step 3: Take the cube root of both sides.

To find $x$, take the cube root of both sides:

$x = \sqrt[3]{64}$ $x = 4$

Solution:

The solution to the equation is:

$x = 4$

Let me know if you'd like a detailed explanation or have any questions!

Here are 5 related questions to expand your understanding:

1. How would the equation change if it were $0 = \sqrt{64 - x^2}$?
2. What are the possible solutions for $x^3 = -64$?
3. How would you solve $0 = \sqrt{81 - x^4}$?
4. How does taking a square root differ from taking a cube root in terms of solutions?
5. What is the difference between real and complex solutions for cube roots?

Tip: When solving equations involving roots, always check for extraneous solutions by substituting back into the original equation.