What are the solutions to the equation x^6-x^3+1=0 ?

The solutions to the equation x^6-x^3+1=0 are

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x^6-x^3+1=0

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

x = cos (\frac{π}{9}) + i sin (\frac{π}{9}), x = cos (\frac{7 π}{9}) + i sin (\frac{7 π}{9}), x = cos (\frac{13 π}{9}) + i sin (\frac{13 π}{9}), x = cos (\frac{π}{9}) - i sin (\frac{π}{9}), x = cos (\frac{5 π}{9}) + i sin (\frac{5 π}{9}), x = cos (\frac{11 π}{9}) + i sin (\frac{11 π}{9})

Solution steps

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{π}{9}) + i sin (\frac{π}{9}), x = cos (\frac{7 π}{9}) + i sin (\frac{7 π}{9}), x = cos (\frac{13 π}{9}) + i sin (\frac{13 π}{9})

Solve

x^{3} = \frac{1}{2} - i \frac{3}{2} : x = cos (\frac{π}{9}) - i sin (\frac{π}{9}), x = cos (\frac{5 π}{9}) + i sin (\frac{5 π}{9}), x = cos (\frac{11 π}{9}) + i sin (\frac{11 π}{9})

The solutions are

x = cos (\frac{π}{9}) + i sin (\frac{π}{9}), x = cos (\frac{7 π}{9}) + i sin (\frac{7 π}{9}), x = cos (\frac{13 π}{9}) + i sin (\frac{13 π}{9}), x = cos (\frac{π}{9}) - i sin (\frac{π}{9}), x = cos (\frac{5 π}{9}) + i sin (\frac{5 π}{9}), x = cos (\frac{11 π}{9}) + i sin (\frac{11 π}{9})

s^{3} + 9 s^{2} + 15 s + 5 = 0

2 x^{3} - 4 x^{2} - 10 x + 20

1.58 = (1 + \frac{x}{2})^{4}

2 x^{3} - 3 x - 12 x = 0

x^{5} + 2 x^{3} + 2 x^{2} = 0

What are the solutions to the equation x^6-x^3+1=0 ?
The solutions to the equation x^6-x^3+1=0 are
Find the zeros of x^6-x^3+1=0
The zeros of x^6-x^3+1=0 are