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

##### 2 Answers

#### Answer:

#### Explanation:

Given:

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

By guessing we can find one solution, namely

#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

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

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

The remaining quadratic

This has discriminant

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

Since

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,

#### Answer:

#### Explanation:

Making

solving