According to the fundamental theorem of Algebra, a polynomial to the degree #n# will have #n# roots. In our case, a cubic will have three roots.
#x=6# is the first solution, which is obtained by solving the equation (taking the cube root of #216#).
If# x=6# is a solution, then #(x-6)# must be a factor of this polynomial. So, if you divide this polynomial by #(x-6)# using algebraic division, you will get the quadratic:
#x^2+6x+36#
Solving this quadratic will give you the remaining two roots:
#x^2+6x+36=0#
#(x+3)^2-9+36=0#
#(x+3)^2+27=0#
#(x+3)= sqrt(-27)#
#(x+3)= ±3sqrt3i#
#x= -3±3sqrt3i#
The catch here is to remember that if you take the square root of any number you get two answers(#±# the answer). I solved this quadratic by 'completing the square', there is also an alternative quadratic formula which can be used to solve quadratics. The non-real solutions exist in conjugate pairs #(a±bi)#.
Therefore, the solutions are:
#x= 6, (-3+sqrt3i), (-3-sqrt3i)#
