Question 73618
The easiest way to do this one is to make this an equation by setting it equal to zero to get:
.
{{{x^3 - 125 = 0}}}
.
Then add 125 to both sides to get:
.
{{{x^3 = 125}}}
.
Then take the cube root of both sides. Note that the cube root of 125 is 5 because 5*5*5 = 125.
.
And the cube root of {{{x^3}}} is just {{{x}}}. So the solution is just:
.
{{{x = 5}}}
. 
Subtract 5 from both sides to get:
.
{{{x-5=0}}} 
.
This tells you that a factor is x-5.  Divide this factor into {{{x^3 - 125}}} and you
find that the quotient is {{{x^2 + 5x + 25}}}
.
This does not factor further into real numbers.  At this stopping point, the answer to
your factoring problem is that it factors into {{{(x-5)*(x^2 + 5x + 25)}}}
.
If you are into complex numbers, you can use the same technique to factor {{{(x^2 + 5x + 25)}}}
Set it equal to zero and apply the quadratic formula to get that the answers are:
.
{{{x = -(5/2)+-i*((5*sqrt(3))/2)}}}
.
These convert into factors of:
.
{{{(x +(5/2)+i*(5*sqrt(3)/2))}}} and {{{(x +(5/2)- i*(5*sqrt(3)/2))}}}
.
As a result this makes your three factors of {{{x^3 - 125}}} equal:
.
{{{(x-5)*(x +(5/2)+i*(5*sqrt(3)/2))*(x +(5/2)- i*(5*sqrt(3)/2))}}}
.
This is tricky to understand so I hope it hasn't overwhelmed you. Give me a response to 
indicate if you need further explanation of this factoring method.