Answer:
The number of complex roots is 6.
Step-by-step explanation:
The given equation is
[tex]x^6+x^3+1=0[/tex]
it can be written as
[tex](x^3)^2+x^3+1=0[/tex]
Substitute [tex]t=x^3[/tex],
[tex](t)^2+t+1=0[/tex]
Using quadratic formula.
[tex]t=\frac{-1\pm \sqrt{1^2-4(1)(1)}}{2}=\frac{-1\pm 3i}{2}[/tex] [tex]t=\frac{-b\pm \sqrt{b^2-4ac}}{2}[/tex]
We know that
[tex]\omega =\frac{-1\pm 3i}{2}[/tex]
It is a complex number.
[tex]x^3=\omega[/tex]
[tex]x=(\omega)^{\frac{1}{3}}[/tex]
Cube root of a complex number is complex.
Therefore the number of complex roots is 6.
