Answer: 6
Explanation:
x⁶ + x³ + 1 = 0
The only possible rational roots are ± 1.
Use synthetic division to see if either root is valid:
1 | 1 0 0 1 0 0 1
| ↓ 1 1 1 2 2 2
1 1 1 2 2 2 3 ← Remainder ≠ 0 so not a factor
-1 | 1 0 0 1 0 0 1
| ↓ -1 1 -1 0 0 0
1 -1 1 0 0 0 1 ← Remainder ≠ 0 so not a factor
Since there are no real roots, all 6 roots must be imaginary (complex). You can graph the equation to confirm (see attached). None of the roots cross the x-axis so they are all imaginary.