Find x-coordinate of stationary point

Find the x-coordinate of the stationary point of the curve $\displaystyle y=\frac{(x-1)^3}{x+1}$, $\displaystyle x>0$

__What I did__

$\displaystyle \frac{dy}{dx}=\frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}$

Stationary Point, dy/dx=0

$\displaystyle \frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}=0$

$\displaystyle (x+1)(3)(x-1)^2-(x-1)^3=0$

$\displaystyle 3(x+1)(x-1)^2=(x-1)^3$

$\displaystyle 3(x+1)=(x-1)$

$\displaystyle 3x+3=x-1$

$\displaystyle 2x+4=0$

$\displaystyle x=-2$

however question states x>0 and ans is 1!