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

1 Answer
Aug 17, 2016

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]}