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

1 Answer
Aug 17, 2016

Answer:

#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#

#color(white)()#
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]}