Solving difference of cubes: what rule requires multiple solutions For my own amusement, I am working through Beginning and Intermediate Algebra by Tyler Wallace.
In Example 483 ( Quadratic In Form ) we arrive at the factors
  $x^3 =1; x^3=8$

Taking the cube root of each side gives us 1;2
Both answers are 'correct' but the answer is not complete. We need to factor the difference in cubes expressions $x^3-1 and x^3-8.
What is the rule that says we have to go on to break down the difference in cubes even though it seems we have an answer to the equation.
For example:
$$x^3=1
\implies x^3-1=0$$
$$\implies (x-1)(x^2+x+1)=0 $$
Clearly, $x=1$ is a root.
The other root lies in the solution to the equation $x^2+x+1 = 0$
Of course, it's discriminant is negative, so it has no real solutions, but you can find complex solutions using the quadratic formula:
$\frac{-1+\sqrt{3}i}{2},\frac{-1-\sqrt{3}i}{2}$
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
We can prove it by expending.
So $$x^3-8=(x-2)(x^2+2x+4)=(x-2)((x+1)^2+3).$$
I hope it what you are looking  for.