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`