What are the solutions to the equation (b^3)/3 =-1/3 ?

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

Find the zeros of (b^3)/3 =-1/3

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

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\frac{b ^{3}}{3} = - \frac{1}{3}

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

Solution steps

\frac{b ^{3}}{3} = - \frac{1}{3}

Multiply both sides by

3

b^{3} = - 1

For

x^{3} = f (a)

the solutions are

Simplify

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

5 x^{3} - 7 x^{2} + 4 x - 8 = 0

2 x^{4} - 4 x^{3} - 4^{2} + 6 x + 9 = 0

x^{3} + 20 x^{2} + 64 x - 100 = 0

- 3 x (36 x^{2} - 25) (x^{2} - 2) = 0

x^{3} - 14 x^{2} = + 56 - 64

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