Question 632223
Using the difference of two cubes where {{{a^3 = (x+5)^3}}} and {{{b^3=27x^3}}}.
{{{a^3-b^3 = 0 }}}
{{{(a-b)(a^2+ab+b^2) = 0}}}
{{{((x+5) - 3x)*((x+5)^2 + 3x(x+5) + (3x)^2) = 0}}}
{{{(-2x + 5)*(x^2 + 10x + 25 + 3x^2 + 15x + 9x^2) = 0}}}
{{{(-2x + 5)*(13x^2 + 25x + 25) = 0}}}
{{{ -2x + 5 = 0}}} or {{{13x^2 + 25x + 25 = 0}}}
Solve: {{{-2x + 5 = 0}}}
{{{highlight(x = 5/2)}}}
Solve: {{{13x^2 + 25x + 25 = 0}}}
*[invoke quadratic "x", 13, 25, 25 ]
There is only one real solution, x = 5/2. 
The other two roots are imaginary numbers as computed above.