We need
#a^3+b^3=(a+b)(a^2-ab+b^2)#
Therefore,
#x^6+64=(x^2)^3+(2^2)^3=(x^2+2^2)(x^4-4x^2+16)#
#{(x^2+4=0),(x^4-4x^2+16=0):}#
#<=>#, #{(x^2=-4),(x^4-4x^2+16=0):}#
#<=>#, #{(x=-2i),(x=2i),(x^2=(4+-(sqrt(16-4*16)))/(2)):}#
#<=>#, #{(x=-2i),(x=2i),(x^2=(4+-(sqrt(-48)))/(2)):}#
#<=>#, #{(x=-2i),(x=2i),(x^2=(2+-(2sqrt3i))):}#
Let #x=a+ib#
Then,
#(a+ib)^2=2+2sqrt3i#
#a^2-b^2+2aib=2+2sqrt3i#
#{(a^2-b^2=2),(2ab=2sqrt3):}#
#{(a^2-b^2=2),(b=sqrt3/a):}#
#a^2-3/a^2=2#
#a^4-2a^2-3=0#
#a^2=(2+-sqrt(4+4*3))/(2)#
#=1+-2#
#a^2={3, -1}#
#a={+-sqrt3, +-i}#
#b={+-, -isqrt3}#
#x={-sqrt3+i, -sqrt3-i}#
Let #x'=a'+ib'#
Then,
#(a'+ib')^2=2-2sqrt3i#
#(a')^2-(b')^2+2(a')(b')i=2-2sqrt3i#
#{((a')^2-(b')^2=2),(2a'b'=-2sqrt3):}#
#{((a')^2-(b')^2=2),(b'=-sqrt3/(a')):}#
#(a')^2-3/(a')^2=2#
#(a')^4-2(a')^2-3=0#
#(a')^2=(2+-sqrt(4+4*3))/(2)#
#=1+-2#
#(a')^2={3, -1}#
#a'={+-sqrt3, +-i}#
#x'={sqrt3-i, sqrt3+i}#
The solutions are #S={2i,-2i,-sqrt3+i, -sqrt3-i, sqrt3-i,sqrt3+i}#
