What are the solutions to the equation j^3=0.125 ?

The solutions to the equation j^3=0.125 are j= 1/2 ,j=-1/4+(sqrt(3))/4 i,j=-1/4-(sqrt(3))/4 i

Find the zeros of j^3=0.125

The zeros of j^3=0.125 are j= 1/2 ,j=-1/4+(sqrt(3))/4 i,j=-1/4-(sqrt(3))/4 i

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j^{3} = 0.125

j = \frac{1}{2}, j = - \frac{1}{4} + \frac{3}{4} i, j = - \frac{1}{4} - \frac{3}{4} i

Solution steps

j^{3} = 0.125

For

x^{3} = f (a)

the solutions are

Simplify

j = \frac{1}{2}, j = - \frac{1}{4} + \frac{3}{4} i, j = - \frac{1}{4} - \frac{3}{4} i

(72) 4 = n^{8}

x, x^{3} - 6 y = 8

9 m^{3} - 54 m^{2} + 72 = 0

(4 - x^{2})^{2} = 0

d^{4} - 81 = 0

What are the solutions to the equation j^3=0.125 ?
The solutions to the equation j^3=0.125 are j= 1/2 ,j=-1/4+(sqrt(3))/4 i,j=-1/4-(sqrt(3))/4 i
Find the zeros of j^3=0.125
The zeros of j^3=0.125 are j= 1/2 ,j=-1/4+(sqrt(3))/4 i,j=-1/4-(sqrt(3))/4 i