What are the solutions to the equation (z^3+i-1)(1-2z+2z^2)=0 ?

The solutions to the equation (z^3+i-1)(1-2z+2z^2)=0 are

Find the zeros of (z^3+i-1)(1-2z+2z^2)=0

The zeros of (z^3+i-1)(1-2z+2z^2)=0 are

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(z^{3} + i - 1) (1 - 2 z + 2 z^{2}) = 0

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}, z = \frac{1}{2} + i \frac{1}{2}, z = \frac{1}{2} - i \frac{1}{2}

Solution steps

(z^{3} + i - 1) (1 - 2 z + 2 z^{2}) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

z^{3} + i - 1 = 0 or 1 - 2 z + 2 z^{2} = 0

Solve

z^{3} + i - 1 = 0 : 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}

Solve

1 - 2 z + 2 z^{2} = 0 : z = \frac{1}{2} + i \frac{1}{2}, z = \frac{1}{2} - i \frac{1}{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}} ( - 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{1}{2} + i \frac{1}{2}, z = \frac{1}{2} - i \frac{1}{2}

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

x^{4} - 144 = 0

2 x^{3} - 14 x^{2} + 14 x + 30 = 0

b^{3} - b = 6

7 x - x^{3} = 0

What are the solutions to the equation (z^3+i-1)(1-2z+2z^2)=0 ?
The solutions to the equation (z^3+i-1)(1-2z+2z^2)=0 are
Find the zeros of (z^3+i-1)(1-2z+2z^2)=0
The zeros of (z^3+i-1)(1-2z+2z^2)=0 are