Question

How do you solve `x^3+1/x^3 = 0` ?

Answer 1

`x_(1,2) = sqrt(3)/2 +- 1/2i`

`x_(3,4) = +-i`

`x_(5,6) = -sqrt(3)/2 +- 1/2i`

Explanation:

Multiply through by `x^3` to find:

`x^6+1 = 0`

For any Real value of `x` we have `x^6 >= 0`, hence `x^6+1 != 0`

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Complex solutions

From de Moivre we have:

`(cos theta + i sin theta)^n = cos n theta + i sin n theta`

Hence Complex solutions:

`x = cos (((2k+1)pi)/6) + i sin (((2k+1)pi)/6)`

That is:

`x_(1,2) = sqrt(3)/2 +- 1/2i`

`x_(3,4) = +-i`

`x_(5,6) = -sqrt(3)/2 +- 1/2i`

These six roots form the vertices of a regular hexagon in the Complex plane.

graph{((x-sqrt(3)/2)^2+(y-1/2)^2-0.002)((x-sqrt(3)/2)^2+(y+1/2)^2-0.002)((x+sqrt(3)/2)^2+(y-1/2)^2-0.002)((x+sqrt(3)/2)^2+(y+1/2)^2-0.002)(x^2+(y-1)^2-0.002)(x^2+(y+1)^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}