What are the solutions to the equation (z+1)^5=(z-1)^5 ?

The solutions to the equation (z+1)^5=(z-1)^5 are

Find the zeros of (z+1)^5=(z-1)^5

The zeros of (z+1)^5=(z-1)^5 are

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(z + 1)^{5} = (z - 1)^{5}

z = \frac{- 5 + 2}{5}, z = - \frac{- 5 + 2}{5}, z = - \frac{5 + 2}{5}, z = - - \frac{5 + 2}{5}

Solution steps

(z + 1)^{5} = (z - 1)^{5}

Expand

(z + 1)^{5} : z^{5} + 5 z^{4} + 10 z^{3} + 10 z^{2} + 5 z + 1

Expand

(z - 1)^{5} : z^{5} - 5 z^{4} + 10 z^{3} - 10 z^{2} + 5 z - 1

z^{5} + 5 z^{4} + 10 z^{3} + 10 z^{2} + 5 z + 1 = z^{5} - 5 z^{4} + 10 z^{3} - 10 z^{2} + 5 z - 1

Subtract

z^{5} - 5 z^{4} + 10 z^{3} - 10 z^{2} + 5 z - 1

from both sides

z^{5} + 5 z^{4} + 10 z^{3} + 10 z^{2} + 5 z + 1 - (z^{5} - 5 z^{4} + 10 z^{3} - 10 z^{2} + 5 z - 1) = z^{5} - 5 z^{4} + 10 z^{3} - 10 z^{2} + 5 z - 1 - (z^{5} - 5 z^{4} + 10 z^{3} - 10 z^{2} + 5 z - 1)

Simplify

10 z^{4} + 20 z^{2} + 2 = 0

Rewrite the equation with

u = z^{2}

and

u^{2} = z^{4}

10 u^{2} + 20 u + 2 = 0

Solve

10 u^{2} + 20 u + 2 = 0 : u = \frac{- 5 + 2 5}{5}, u = - \frac{5 + 2}{5}

u = \frac{- 5 + 2 5}{5}, u = - \frac{5 + 2}{5}

Substitute back

u = z^{2},

solve for

z

Solve

z^{2} = \frac{- 5 + 2 5}{5} : z = \frac{- 5 + 2}{5}, z = - \frac{- 5 + 2}{5}

Solve

z^{2} = - \frac{5 + 2}{5} : z = - \frac{5 + 2}{5}, z = - - \frac{5 + 2}{5}

The solutions are

z = \frac{- 5 + 2}{5}, z = - \frac{- 5 + 2}{5}, z = - \frac{5 + 2}{5}, z = - - \frac{5 + 2}{5}

x^{3} - 12 x^{2} + 48 x - 64 = 0

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

m^{3} - 3 m^{2} + 4 = 0

x^{3} = 1

x^{4} + 1 = 0

What are the solutions to the equation (z+1)^5=(z-1)^5 ?
The solutions to the equation (z+1)^5=(z-1)^5 are
Find the zeros of (z+1)^5=(z-1)^5
The zeros of (z+1)^5=(z-1)^5 are