I substituted a=x3 which gave me a^2 -a -2 = 0, solving for a you get (a-2)(a+1)=0 so a=2 or a=-1 substitue x^3 for a and you get directly to x=2^(1/3) or x=-1
Yes, there are 6 roots, two of which are real, and four are complex. It is obvious that x^6-x^3-2=(x+1)(x^3-2)(x^2-x+1): the first factor from the observation that x=-1 is a solution, and that if x^3=2 then x is a solution. The third factor can be obtained by dividing x^6-x^3-2 by the first two factors. From the first factor we have the solutions -1, from the second factors we have 2^(1/3) and the two complex solutions -2**(1/3)/2 +2^**(1/3)*sqrt(3)*i/2 and -2**(1/3)/2 - 2**(1/3)*sqrt(3)*i/2, and finally x=1/2 + sqrt(3)*i/2 and x=1/2 - sqrt(3)*i/2 from the third factor. This is all really very straightforward.
