Prove that when using the quadratic formula for the two complex solutions, the radicand in the final answer is always 3 (sqrt3). (Sums and differences of cubes)

This is not a very clear question, but it seems that you need to show that the complex solutions of the equation of the form
x^3 + a^3 = 0 or x^3 - a^3 = 0   will contain a radical of 3 (sqrt(3) ). Here, a is assumed to be a positive real number.
This can be shown by factoring the left side using the formula for the sum of cubes (or difference of cubes, for the second equation):
(x + a)(x^2 -ax + a^2) = 0
This is satisfied when x = -a (this is the real solution) or when x^2 - ax + a^2 = 0 . The quadratic equation can be solved by using the formula:
x = (a +-sqrt(a^2 - 4a^2))/2 = (a +-isqrt(3)a)/2
These are two complex solutions, as the radical of the negative number -3a^2 is an imaginary number isqrt(3)a . No matter what the value of a is, the equation of the form shown above will always have two complex solutions which contain the square root of 3, and one real solution.
The same can be shown for the equation of the form x^3 - a^3 , except the real solution will be x = a and the complex solutions will be
x = (-a +-isqrt(3)a)/2 .