Question 393464
Facor:
{{{27x^3+125 = 0}}} Did you notice that you have the sum of two cubes?
{{{(3x)^3+(5)^3 = 0}}}
The sum of two cubes can be factored thus:
{{{A^3+B^3 = (A+B)(A^2-AB+B^2)}}}
In your problem, A = 3x and B = 5, so...
{{{27x^3+125 = 0}}} becomes:
{{{(3x+5)(9x^2-15x+25) = 0}}} Apply the zero product rule:
{{{3x+5 = 0}}} or {{{9x^2-15x+25 = 0}}} 
Do the first factor:
{{{3x+5 = 0}}} Subtract 5.
{{{3x = -5}}} Divide by 3.
{{{x[1] = -5/3}}}
Do the second factor:
{{{9x^2-15x+25 = 0}}} Use the quadratic formula to solve: {{{x = (-b+-sqrt(b^2-4ac))/2a}}} and a = 9, b = -15, and c = 25.
{{{x = (-(-15)+-sqrt((-15)^2-4(9)(25)))/2(9)}}}
{{{x = (15+-sqrt(225-900))/18}}}
{{{x = (15+-sqrt(-675))/18}}} and when you work this out, you should get:
{{{x[2] = 0.833+1.44i}}} and {{{x[3] = 0.833-1.44i}}}