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^6+1 = 0
For any Real value of
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]}