解：设复数为z=x+yi(x，y为实数)有(x+yi)³=8，得：x³+3x²yi-3xy²-y³i=8，则有x³-3xy²=8，3x²y-y³=0；x=-1，y=±√3；z=-1+√3i或z=-1-√3i
z^3-8=0z^3=8= 8[cos(2kπ) + isin(2kπ) ]z = 2√2[cos(2kπ/3) + isin(2kπ/3) ] ; k=0,1,2
