Question 155348
Notice that your equation:
{{{x^3-8=0}}}
is a "difference of two cubes"
That is, both terms have cube roots:
the cube root of x^3 is x
the cube root of 8^3 is 2
.
This site describes these "special cases":
http://www.purplemath.com/modules/specfact2.htm
.
Bottom-line:
If you see a situation of a "difference of two cubes", you can factor thus:
a^3 – b^3 = (a – b)(a^2 + ab + b^2)
.
In our case:
a is x
b is 2
.
Therefore, we can rewrite:
{{{x^3-8=0}}}
as
{{{(x-2)(x^2 + 2x + 2^2)=0}}}
{{{(x-2)(x^2 + 2x + 4)=0}}}
.
Finally, to solve, we set each factor on the left to zero:
First term:
(x-2) = 0
x = 2 (Here's one "real" solution)
.
Second term:
(x^2 + 2x + 4)=0
Here's where we need to apply the "quadratic equation":
Note: you will find that there are no real solutions, only 2 imaginary ones:

*[invoke quadratic "x", 1, 2, 4 ]