Solving $k(k+1)(k-1)=k$

Question

The function $y=e^{kx}$ satisfies the equation

$$\left(\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}\right)\left(\frac{dy}{dx}-y\right)=y\frac{dy}{dx}$$

I found the derivative and the second derivative of $y$, resulting in

$$\left(k^{2}e^{kx}+ke^{kx}\right)\left(ke^{kx}-e^{kx}\right)=ke^{2kx}$$

Dividing by $e^{2kx}$:

$$k(k+1)(k-1)=k \tag{$\star$}$$ I don't understand this part. How do you simplify $(\star)$ to find the values of $k$?

I know one of the roots is $0$, but not the other two.

Answer 1

$k(k+1)(k-1)=k$

$k(k+1)(k-1) - k = 0$

$k(k^2-1) - k = 0$

$k(k^2-1 - 1)= 0$

$k(k^2 - 2)=0$

$k =0$ or $k^2 = 2 \implies k = \pm\sqrt 2$

$\therefore k = 0, - \sqrt 2, \sqrt 2$