How do you solve #root3(x)=x-6#?

Given:

#root(3)(x) = x-6#

By guessing we can find one solution, namely #x=8# since:

#root(3)(8) = root(3)(2^3) = 2 = 8-6#

Let us see if we can find some more.

Cube both sides of the given equation to get:

#x=(x-6)^3 = x^3-18x^2+108x-216#

Subtract #x# from both sides to get:

#0 = x^3-18x^2+107x-216#

#color(white)(0) = (x-8)(x^2-10x+27)#

The remaining quadratic #x^2-10x+27# is in the form #ax^2+bx+c# with #a=1#, #b=-10# and #c=27#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(-10))^2-4(color(blue)(1))(color(blue)(27)) = 100-108 = -8#

Since #Delta < 0#, this quadratic has non-real Complex zeros.

If we want we can find them using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (10+-sqrt(-8))/2#

#color(white)(x) = 5+-sqrt(2)i#

Interestingly, neither of these is a solution of the original equation, since the principal cube root of both of these is not the one you need to satisfy the equation.

For example, #5+sqrt(2)i# is in Q1, with principal cube root in Q1, but #5+sqrt(2)i - 6# is in Q2.

Making #y = root(3)(x)# we have

#y^3-y=6# or

#(y-1)y(y+1) = 6# and we have that #y=2# is a root then

#(y-2)(y^2+ay+b)+c=y^3-y-6# then

#{(6 - 2 b + c = 0),(1 - 2 a + b = 0),(a - 2= 0):}#

solving

#a=2, b=3, c=0# then

#(y-2)(y^2+2y+3)=y^3-y-6# but the roots of #y^2+2y+3# are complex conjugate so the only real root is

#y=root(3)(x)=2 rArr x=8#