Please solve this equation. As you write down the steps, please give me a description on what you did.

y^3 - 16y^1/2 + 64 = 0

LOL

y must be positive if all is real numbers
But for any +y, the answer is >0

I mean the left side is greater than zero

To solve the equation y^3 - 16y^(1/2) + 64 = 0, we can use a method called factoring. The goal is to express the equation as a product of two factors, so that we can set each factor equal to zero and solve for y.

Step 1: Look for any common factors. In this case, the equation does not have any common factors.

Step 2: Notice that the equation can be rewritten as (y^(3/2))^2 - 2 * (y^(3/2)) * 4 + 4^2 = 0.

Step 3: Apply the concept of a perfect square trinomial. The expression above can be factored as (y^(3/2) - 4)^2 = 0.

Step 4: Set each factor equal to zero and solve for y:
y^(3/2) - 4 = 0

Step 5: Add 4 to both sides of the equation:
y^(3/2) = 4

Step 6: Square both sides of the equation to eliminate the exponent:
(y^(3/2))^2 = 4^2
y^3 = 16

Step 7: Take the cube root of both sides to isolate y:
y = ∛16

Step 8: Simplify the cube root of 16:
y = 2

After going through these steps, we have found that the solution to the equation y^3 - 16y^(1/2) + 64 = 0 is y = 2.