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

Jul 19, 2017

$x = 8$

#### Explanation:

Given:

$\sqrt[3]{x} = x - 6$

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

$\sqrt[3]{8} = \sqrt[3]{{2}^{3}} = 2 = 8 - 6$

Let us see if we can find some more.

Cube both sides of the given equation to get:

$x = {\left(x - 6\right)}^{3} = {x}^{3} - 18 {x}^{2} + 108 x - 216$

Subtract $x$ from both sides to get:

$0 = {x}^{3} - 18 {x}^{2} + 107 x - 216$

$\textcolor{w h i t e}{0} = \left(x - 8\right) \left({x}^{2} - 10 x + 27\right)$

The remaining quadratic ${x}^{2} - 10 x + 27$ is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 10$ and $c = 27$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{- 10}\right)}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{27}\right) = 100 - 108 = - 8$

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

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{10 \pm \sqrt{- 8}}{2}$

$\textcolor{w h i t e}{x} = 5 \pm \sqrt{2} i$

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.

Jul 20, 2017

$x = 8$

#### Explanation:

Making $y = \sqrt[3]{x}$ we have

${y}^{3} - y = 6$ or

$\left(y - 1\right) y \left(y + 1\right) = 6$ and we have that $y = 2$ is a root then

$\left(y - 2\right) \left({y}^{2} + a y + b\right) + c = {y}^{3} - y - 6$ then

$\left\{\begin{matrix}6 - 2 b + c = 0 \\ 1 - 2 a + b = 0 \\ a - 2 = 0\end{matrix}\right.$

solving

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

$\left(y - 2\right) \left({y}^{2} + 2 y + 3\right) = {y}^{3} - y - 6$ but the roots of ${y}^{2} + 2 y + 3$ are complex conjugate so the only real root is

$y = \sqrt[3]{x} = 2 \Rightarrow x = 8$