What are the solutions to the equation ((x+1)/2)^3=(-1)/8 ?

The solutions to the equation ((x+1)/2)^3=(-1)/8 are x=-2,x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2

Find the zeros of ((x+1)/2)^3=(-1)/8

The zeros of ((x+1)/2)^3=(-1)/8 are x=-2,x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2

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(\frac{x + 1}{2})^{3} = \frac{- 1}{8}

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

Solution steps

(\frac{x + 1}{2})^{3} = \frac{- 1}{8}

Multiply both sides by

8

(x + 1)^{3} = - 1

Expand

(x + 1)^{3} : x^{3} + 3 x^{2} + 3 x + 1

x^{3} + 3 x^{2} + 3 x + 1 = - 1

Move

1

to the left side

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

Factor

x^{3} + 3 x^{2} + 3 x + 2 : (x + 2) (x^{2} + x + 1)

(x + 2) (x^{2} + x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x^{2} + x + 1 = 0

Solve

x + 2 = 0 : x = - 2

Solve

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

The solutions are

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

r^{3} + 10 r^{2} + 33 r - 29 = 0

1.477 = (1 + g)^{5}

(x + 2) (2 x^{2} - x - 1) = 0

320 = 62500 \cdot (\frac{1 - r}{100})^{3}

x^{4} + 4 x^{3} + 8 x^{2} + 8 x + 4 = 0

What are the solutions to the equation ((x+1)/2)^3=(-1)/8 ?
The solutions to the equation ((x+1)/2)^3=(-1)/8 are x=-2,x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2
Find the zeros of ((x+1)/2)^3=(-1)/8
The zeros of ((x+1)/2)^3=(-1)/8 are x=-2,x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2