system

To find the possible values of ab, a+b, a, and b, we'll use the given equations and solve them simultaneously.

(a) Finding all possible values of ab:

From the equation a - b = 4, we can rewrite it as a = b + 4.

Substituting this value of a into the equation a^3 - b^3 = 0, we get:
(b + 4)^3 - b^3 = 0

Expanding the equation, we have:
(b^3 + 12b^2 + 48b + 64) - b^3 = 0

Simplifying the equation, we get:
12b^2 + 48b + 64 = 0

Dividing the equation by 4 to simplify it further, we have:
3b^2 + 12b + 16 = 0

Using the quadratic formula, we can solve for b:
b = (-12 ± √(12^2 - 4316))/(2*3)
b = (-12 ± √(144 - 192))/(6)
b = (-12 ± √(-48))/(6)

Since the discriminant is negative, there are no real solutions for b. Therefore, there are no possible real values for ab.

(b) Finding all possible values of a + b:

Given a - b = 4, we can rewrite it as a = b + 4.

Substituting this value of a into the equation a + b, we get:
(b + 4) + b = 2b + 4

So, the possible values of a + b are all real numbers of the form 2b + 4.

(c) Finding all possible values of a and b:

We have a - b = 4. By substituting the value of a from this equation into the equation a + b = 2b + 4, we get:
(b + 4) + b = 2b + 4

Simplifying the equation, we have:
2b + 4 = 2b + 4

This equation is true for all values of b. Therefore, there are infinitely many possible values for a and b that satisfy the given conditions.

In summary:
(a) There are no possible real values for ab.
(b) The possible values of a + b are all real numbers of the form 2b + 4.
(c) There are infinitely many possible values for a and b that satisfy the given conditions.