Question 362294
{{{x^6-65x^3+64=(x^3-1)(x^3-64) = (x-1)(x^2+x+1)(x-4)(x^2+4x+16)=0}}}. The real roots are 1 or 4.  The complex roots are gotten using the quadratic formula:
{{{x = (-1 +- sqrt(3)i)/2}}} and {{{x = -2 +- 2sqrt(3 )i}}}