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

The solutions to the equation (1+y^3)^2=0 are y=-1,y= 1/2-(sqrt(3))/2 i,y= 1/2+(sqrt(3))/2 i

Find the zeros of (1+y^3)^2=0

The zeros of (1+y^3)^2=0 are y=-1,y= 1/2-(sqrt(3))/2 i,y= 1/2+(sqrt(3))/2 i

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(1 + y^{3})^{2} = 0

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

Solution steps

(1 + y^{3})^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

1 + y^{3} = 0 : y = - 1, y = \frac{1}{2} - \frac{3}{2} i, y = \frac{1}{2} + \frac{3}{2} i

The solutions are

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

2 = x^{3}

3 = y^{5} + y^{3} + y

9 x^{3} - 45 x^{2} - 52 x - 12 = 0

z^{6} + 64 = 0

y^{3} - 9 y - 12 = 0

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