Cube root of a number is equal to its own cube. What are all the possibilities for this number?

Question

Isaiah Anthony · Accepted Answer

At least #0, -1# and #1#.

With #Arg(z) in (-pi, pi]# there are additional Complex solutions:

#-sqrt(2)/2+-sqrt(2)/2i#

Given: #x^3=root(3)(x)# Cubing both sides we get: #x^9 = x# Note that cubing both sides may (and actually does) introduce spurious solutions. One solution is #x=0#, which is a solution of the original equation. If #x != 0# then we can divide both sides by #x# to get: #x^8 = 1# This has Real solutions #x = +-1#, both of which are solutions of the original problem, if we follow the normal convention that #root(3)(1) = 1# and #root(3)(-1) = -1#. #color(white)()# Any possible Complex solutions of this octic equation can be expressed in the form: #cos((k pi)/4) + i sin((k pi)/4)# for #k = 0, 1, 2, 3, 4, 5, 6, 7# That is: #x = { (1), (sqrt(2)/2+sqrt(2)/2i), (i), (-sqrt(2)/2+sqrt(2)/2i), (-1), (-sqrt(2)/2-sqrt(2)/2i), (-i), (sqrt(2)/2-sqrt(2)/2i) :}# The question as to whether any of these is a solution of the original equation depends on what you mean by #root(3)(x)# if #x in CC#. Suppose, consistent with #Arg(z) in (-pi, pi]#, we define: #root(3)(cos theta + i sin theta) = cos (theta/3) + i sin (theta/3)# when #-pi < theta < pi# Most of the solutions of #x^8=1# are not solutions of #x^3 = root(3)(x)#, but the following two are: #x = { (-sqrt(2)/2+sqrt(2)/2i), (-sqrt(2)/2-sqrt(2)/2i) :}# #color(white)()# Note that if you prefer #Arg(z) in [0, 2pi)# then the valid solutions apart from #0, 1 and -1# are: #x = { (-sqrt(2)/2+sqrt(2)/2i), (-i) :}#

Abigail Allen · Answer

is there a more systematic way?

#x^8 = e^(2nipi)# #x = e^(ni pi/4)# where #n=1,8#

Eva Adamson · Answer

The solutions are #+-1. +-i, (+-1+-i)/sqrt2#

Let the number be x.

#x^(1/3)=x^3#. #x^(3-1/3)=1# #x^(8/3)=1# # x=1^(3/8). =(1^3)^(1/8)=1^(1/8)#

#So, x=(e^(i2n pi))^(1/8)=e^((i2npi)/8)#, n = any integer, including 0.

The values for n = 0, 1, 2.., 7 are repeated, in a cycle, for other values of n.

The eight distinct values are determinate and are listed in the answer.

Importantly, observe that x = 0 is not a solution. .

Isaiah Anthony · Answer

undefined The only number that satisfies the condition "the cube root of a number is equal to its own cube" is 1.