a^6 - 9a^3 + 8 = 0 factoring directly, we have
(a^3 - 8) (a^3 -1) = 0
Factoring these as differences of cubes, we have
(a - 2)(a^2 + 2a + 4)(a - 1)(a^2 + a + 1) = 0
Setting the first and third linear factors to 0 produces two real solutions....a = 2 and a = 1
Setting the second factor to 0 and using thequadratic formula produces two complex solutions :
a = [ -2 ± √ ( 4 - 16) / 2 = [ -2 ± √ [( - 12)] / 2 = [ -2 ± 2√ ( -3 ) ] / 2 = -1 ± i√3
Setting the fourth factor to 0 and using thequadratic formula produces the two remaining complex solutions :
a = [-1 ± √ (1 - 4) ] / 2 = [ -1 ± √ (-3) ] / 2 = [ -1 ± i√ (3) ] / 2