Question 422795
Let {{{z = x^3}}}. Our polynomial becomes


{{{z^2 - 19z = 216}}}

{{{z^2 - 19z - 216 = 0}}}. We can factor this:

{{{(z-27)(z+8) = 0}}}

{{{z = 27}}} or {{{z = -8}}}

Hence, all solutions x must satisfy {{{x^3 = 27}}} or {{{x^3 = -8}}}.

If you know roots of unity, we can conclude that the roots of {{{x^3 = 27}}} are evenly spaced on the complex plane starting with x = 3. The other two roots are {{{3(cos(2pi/3) + i*sin(2pi/3))}}} and {{{3(cos(4pi/3) + i*sin(4pi/3))}}}. For {{{x^3 = -8}}} we have {{{-2}}}, {{{2(cos(5pi/3) + i*sin(5pi/3))}}} and {{{2(cos(pi/3) + i*sin(pi/3))}}} as our roots for x.