What are the solutions to the equation z^3=(z-1)^3 ?

The solutions to the equation z^3=(z-1)^3 are z= 1/2-i(sqrt(3))/6 ,z= 1/2+i(sqrt(3))/6

Find the zeros of z^3=(z-1)^3

The zeros of z^3=(z-1)^3 are z= 1/2-i(sqrt(3))/6 ,z= 1/2+i(sqrt(3))/6

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z^{3} = (z - 1)^{3}

z = \frac{1}{2} - i \frac{3}{6}, z = \frac{1}{2} + i \frac{3}{6}

Solution steps

z^{3} = (z - 1)^{3}

Expand

(z - 1)^{3} : z^{3} - 3 z^{2} + 3 z - 1

z^{3} = z^{3} - 3 z^{2} + 3 z - 1

Switch sides

z^{3} - 3 z^{2} + 3 z - 1 = z^{3}

Move

z^{3}

to the left side

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

Solve with the quadratic formula

z_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 3 ) ( - 1 )}{2 ( - 3 )}

Simplify

3^{2} - 4 (- 3) (- 1) : 3 i

z_{1, 2} = \frac{- 3 \pm 3 i}{2 ( - 3 )}

Separate the solutions

z_{1} = \frac{- 3 + 3 i}{2 ( - 3 )}, z_{2} = \frac{- 3 - 3 i}{2 ( - 3 )}

z = \frac{- 3 + 3 i}{2 ( - 3 )} : \frac{1}{2} - i \frac{3}{6}

z = \frac{- 3 - 3 i}{2 ( - 3 )} : \frac{1}{2} + i \frac{3}{6}

The solutions to the quadratic equation are:

z = \frac{1}{2} - i \frac{3}{6}, z = \frac{1}{2} + i \frac{3}{6}

10 x^{2} - 20 x + 1 = 0

x^{2} - 25 x + 36 = 0

5 x^{2} - 7 x = - 4 x + 3 x^{2}

x^{2} + 9 = 4 x

- 6 m^{2} - 6 = - 7 m - 9

What are the solutions to the equation z^3=(z-1)^3 ?
The solutions to the equation z^3=(z-1)^3 are z= 1/2-i(sqrt(3))/6 ,z= 1/2+i(sqrt(3))/6
Find the zeros of z^3=(z-1)^3
The zeros of z^3=(z-1)^3 are z= 1/2-i(sqrt(3))/6 ,z= 1/2+i(sqrt(3))/6