What are the solutions to the equation w^3=-125 ?

The solutions to the equation w^3=-125 are w=-5,w= 5/2-(5sqrt(3))/2 i,w= 5/2+(5sqrt(3))/2 i

Find the zeros of w^3=-125

The zeros of w^3=-125 are w=-5,w= 5/2-(5sqrt(3))/2 i,w= 5/2+(5sqrt(3))/2 i

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w^{3} = - 125

w = - 5, w = \frac{5}{2} - \frac{5 3}{2} i, w = \frac{5}{2} + \frac{5 3}{2} i

Solution steps

w^{3} = - 125

For

x^{3} = f (a)

the solutions are

Simplify

w = - 5, w = \frac{5}{2} - \frac{5 3}{2} i, w = \frac{5}{2} + \frac{5 3}{2} i

(x^{2} - 49) (x^{2} + 16) = 0

x^{3} - 4 x^{2} - 12 x = 0

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

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

7 x^{2} (3 x - 5) = 0

What are the solutions to the equation w^3=-125 ?
The solutions to the equation w^3=-125 are w=-5,w= 5/2-(5sqrt(3))/2 i,w= 5/2+(5sqrt(3))/2 i
Find the zeros of w^3=-125
The zeros of w^3=-125 are w=-5,w= 5/2-(5sqrt(3))/2 i,w= 5/2+(5sqrt(3))/2 i