Algebra

(a) To find the possible values of ab, we can use the identity a^3 - b^3 = (a - b)(a^2 + ab + b^2) and substitute a - b = 1 and a^3 - b^3 = 1 to get:

1 = (a - b)(a^2 + ab + b^2) = (1)(a^2 + ab + b^2) = a^2 + ab + b^2

Now, we can use the identity (a + b)^2 = a^2 + 2ab + b^2 and substitute a - b = 1 to get:

(a + b)^2 = (a - b)^2 + 4ab = 1 + 4ab

So, ab = ((a + b)^2 - 1)/4. Therefore, ab can take any value greater than or equal to -1/4.

(b) To find the possible values of a+b, we can add a - b = 1 to a^3 - b^3 = 1 to get:

a^3 + a - b^3 - b = 2a

Now, we can use the identity a^3 - b^3 = (a - b)(a^2 + ab + b^2) and substitute a - b = 1 to get:

a^2 + ab + b^2 = 1

So, a+b = 2a - 1 = 2b + 1. Therefore, a+b can take any value.

(c) To find the possible values of a and b, we can solve the system of equations:  ACEFlare

a - b = 1 a^3 - b^3 = 1

Substituting a - b = 1 into a^3 - b^3 = 1 gives:

(a - b)(a^2 + ab + b^2) = 1

Using a^2 + ab + b^2 = 1, we have:

a - b = 1 a^2 + ab + b^2 = 1

Substituting a = 1 + b into a^2 + ab + b^2 = 1 gives:

3b^2 + 3b = 0

So, b = 0 or b = -1. If b = 0, then a = 1, and if b = -1, then a = 0. Therefore, the possible values of a and b are (1, 0) and (0, -1).