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.
 A: 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: The 'Rule' is called the Fundamental Theorem of Algebra, which states that every (non-zero, with complex coefficients) polynomial (in one variable) of degree $n$ must have $n$ roots in $\mathbb C$, including multiplicities (e.g. $(x-1)^2$ has a root of $x=1$ with multiplicity $2$.)
This rule shows that there could be roots we are missing if we simply find the obvious roots of such polynomials.
A: In this example the author isn't looking for a solution, but for all solutions.
The solutions $x=1$ and $x=2$ are the only solutions in the real numbers: If you use the fact that the function $f(x)=x^3$ is increasing, then it follows that there can be only one solution to $x^3=1$, and to $x^3=8$, and indeed you are done when you have found $x=1$ and $x=2$.
But over the complex numbers this argument doesn't hold. There could be more solutions. The factorization
$$x^3-1=(x-1)(x^2+x+1),$$
helps you to find them; either $x=1$ or $x^2+x+1=0$. Then the quadratic formula tells you precisely which two complex numbers satisfy $x^2+x+1=0$. The same goes for $x^3-8$.
A: $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
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.
