What are the solutions to the equation k^3=27 ?

The solutions to the equation k^3=27 are k=3,k=-3/2+(3sqrt(3))/2 i,k=-3/2-(3sqrt(3))/2 i

Find the zeros of k^3=27

The zeros of k^3=27 are k=3,k=-3/2+(3sqrt(3))/2 i,k=-3/2-(3sqrt(3))/2 i

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k^{3} = 27

k = 3, k = - \frac{3}{2} + \frac{3 3}{2} i, k = - \frac{3}{2} - \frac{3 3}{2} i

Solution steps

k^{3} = 27

For

x^{3} = f (a)

the solutions are

Simplify

k = 3, k = - \frac{3}{2} + \frac{3 3}{2} i, k = - \frac{3}{2} - \frac{3 3}{2} i

n^{3} - 8 = - 2

x, x^{3} - y^{2} = 3 + 5 y

x^{4} + 10 x^{3} + 35 x^{2} + 50 x + 24 = 0

- x^{3} - x^{2} + 6 x = 0

2 (x + 1)^{3} - 4 = 12

What are the solutions to the equation k^3=27 ?
The solutions to the equation k^3=27 are k=3,k=-3/2+(3sqrt(3))/2 i,k=-3/2-(3sqrt(3))/2 i
Find the zeros of k^3=27
The zeros of k^3=27 are k=3,k=-3/2+(3sqrt(3))/2 i,k=-3/2-(3sqrt(3))/2 i