What are the solutions to the equation y^3=1 ?

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

Find the zeros of y^3=1

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

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y^{3} = 1

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

Solution steps

y^{3} = 1

For

x^{3} = f (a)

the solutions are

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

Simplify

\frac{- 1 + 3 i}{2} : - \frac{1}{2} + \frac{3}{2} i

Simplify

\frac{- 1 - 3 i}{2} : - \frac{1}{2} - \frac{3}{2} i

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

(\frac{x + 1}{2}) \cdot x^{3} + 5 = - 22

25 y^{3} - 36 y = 0

2 = (1 + x)^{10}

n^{3} - n = 720

x^{3} - 9 x^{2} + 3 x + 45 = 0

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