What are the solutions to the equation (x+1)^3=x^3-5 ?

The solutions to the equation (x+1)^3=x^3-5 are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2

Find the zeros of (x+1)^3=x^3-5

The zeros of (x+1)^3=x^3-5 are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2

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(x + 1)^{3} = x^{3} - 5

x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

Solution steps

(x + 1)^{3} = x^{3} - 5

Expand

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

x^{3} + 3 x^{2} + 3 x + 1 = x^{3} - 5

Move

5

to the left side

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

Move

x^{3}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

Simplify

3^{2} - 4 \cdot 3 \cdot 6 : 37 i

x_{1, 2} = \frac{- 3 \pm 3 7 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 3 + 3 7 i}{2 \cdot 3}, x_{2} = \frac{- 3 - 3 7 i}{2 \cdot 3}

x = \frac{- 3 + 3 7 i}{2 \cdot 3} : - \frac{1}{2} + i \frac{7}{2}

x = \frac{- 3 - 3 7 i}{2 \cdot 3} : - \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{7}{2}, x = - \frac{1}{2} - i \frac{7}{2}

(\frac{x - 1}{2})^{2} = 0

x, 4 x^{2} \cdot 4 y^{2} = 12

(x - 3) (x + 1) = (3 x + 1) (2 x + 1)

h, h + s^{2} = h^{2} s

43 \cdot x^{2} + 5 x - 23 = 0, x

What are the solutions to the equation (x+1)^3=x^3-5 ?
The solutions to the equation (x+1)^3=x^3-5 are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2
Find the zeros of (x+1)^3=x^3-5
The zeros of (x+1)^3=x^3-5 are x=-1/2+i(sqrt(7))/2 ,x=-1/2-i(sqrt(7))/2