What are the solutions to the equation roots z^6-2z^3+2 ?

The solutions to the equation roots z^6-2z^3+2 are

Find the zeros of roots z^6-2z^3+2

The zeros of roots z^6-2z^3+2 are

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roots

z^{6} - 2 z^{3} + 2

z = \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{4}, z = \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}

Solution steps

z^{6} - 2 z^{3} + 2

The roots are the intercepts with the x-axis

(y = 0)

z^{6} - 2 z^{3} + 2 = 0

Rewrite the equation with

u = z^{3}

and

u^{2} = z^{6}

u^{2} - 2 u + 2 = 0

Solve

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

u = 1 + i, u = 1 - i

Substitute back

u = z^{3},

solve for

z

Solve

z^{3} = 1 + i : z = \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{4}

Solve

z^{3} = 1 - i : z = \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}

The solutions are

z = \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{4}, z = \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}

0 = x^{3} + 21 x - 90

(9 x^{3} + 6 x^{2} - 48 x + 10) = - 10

624 = 300 x + 3 x^{3}

(x - 1)^{3} = 8

x^{3} - 15 x^{2} + 63 x - 49 = 0

What are the solutions to the equation roots z^6-2z^3+2 ?
The solutions to the equation roots z^6-2z^3+2 are
Find the zeros of roots z^6-2z^3+2
The zeros of roots z^6-2z^3+2 are