solving 3x(x^3-1) = 0 correctly for terms?

I must have forgot how to do this type of problem:

When I solve 3x(x^3-1) = 0 for x with my calculator's SOLVE, for x, I get x=0 and x=1.

Is this because

x^3 = 0/3x+1 = 0+1 = 1 ?
and
3x = 0/(x^3-1)
x = 0/x(x^3-1)
= 0/(x^2-x) = 0 ?

How do you get X=1 & X=0 ?


thanks

3x(x^3-1) = 0
x^3 - 1 = 0

use the rational roots test to find a root of (x^3-1) we found that 1 is a zero, so (x-1) is a root

now dividing (x^3-1) by (x-1) you get

x^2+x+1

solving that polynomial, you will get two complex roots (non-real soultions)

thus you are left with:
3x = 0
(x-1) = 0


solving for each, you get x = 0 and x = 1

Take a look at the following expression: cab, where c is not zero.

For cab to be 0, what must be true?