What are the solutions to the equation roots x^6+x^3+1 ?

The solutions to the equation roots x^6+x^3+1 are

Find the zeros of roots x^6+x^3+1

The zeros of roots x^6+x^3+1 are

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roots

x^{6} + x^{3} + 1

x = cos (\frac{2 π}{9}) + i sin (\frac{2 π}{9}), x = cos (\frac{8 π}{9}) + i sin (\frac{8 π}{9}), x = cos (\frac{14 π}{9}) + i sin (\frac{14 π}{9}), x = cos (\frac{2 π}{9}) - i sin (\frac{2 π}{9}), x = cos (\frac{4 π}{9}) + i sin (\frac{4 π}{9}), x = cos (\frac{10 π}{9}) + i sin (\frac{10 π}{9})

Solution steps

x^{6} + x^{3} + 1

The roots are the intercepts with the x-axis

(y = 0)

x^{6} + x^{3} + 1 = 0

Rewrite the equation with

u = x^{3}

and

u^{2} = x^{6}

u^{2} + u + 1 = 0

Solve

u^{2} + u + 1 = 0 : u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Substitute back

u = x^{3},

solve for

x

Solve

x^{3} = - \frac{1}{2} + i \frac{3}{2} : x = cos (\frac{2 π}{9}) + i sin (\frac{2 π}{9}), x = cos (\frac{8 π}{9}) + i sin (\frac{8 π}{9}), x = cos (\frac{14 π}{9}) + i sin (\frac{14 π}{9})

Solve

x^{3} = - \frac{1}{2} - i \frac{3}{2} : x = cos (\frac{2 π}{9}) - i sin (\frac{2 π}{9}), x = cos (\frac{4 π}{9}) + i sin (\frac{4 π}{9}), x = cos (\frac{10 π}{9}) + i sin (\frac{10 π}{9})

The solutions are

x = cos (\frac{2 π}{9}) + i sin (\frac{2 π}{9}), x = cos (\frac{8 π}{9}) + i sin (\frac{8 π}{9}), x = cos (\frac{14 π}{9}) + i sin (\frac{14 π}{9}), x = cos (\frac{2 π}{9}) - i sin (\frac{2 π}{9}), x = cos (\frac{4 π}{9}) + i sin (\frac{4 π}{9}), x = cos (\frac{10 π}{9}) + i sin (\frac{10 π}{9})

32 xxx - 234 xx + 410 x - 39 = 0

x^{3} - 6 x^{2} + 11 x - 6 = 0

z^{5} - z = 0

1 + z^{4} = 0

x^{3} + 8 x^{2} = 0

What are the solutions to the equation roots x^6+x^3+1 ?
The solutions to the equation roots x^6+x^3+1 are
Find the zeros of roots x^6+x^3+1
The zeros of roots x^6+x^3+1 are