Obviously $ -2 \le x$ ; we will prove that $ x \le 2$ Indeed ; with $ 2< x $ ; we have : $ x^3 - 3x = x(x^2 - 4) + x > x > \sqrt{2x} > \sqrt{x+2} $ So ; we only need to consider the case $ -2 \le x \le 2$ But with the above condition ; we can setting $ x = 2cos y \ \ ; 0 \le y \le \pi$ $\implies \sqrt{x+2} = \sqrt{2( 1 + cos y )} = \sqrt{4 cos^2 \frac{y}{2} } = 2cos \frac{y}{2}$ ( $ 0 \le y \le \pi \implies 0 \le \frac{y}{2} \le \frac{\pi}{2} \implies 0 \le cos \frac{y}{2}$ ) and $ x^3 - 3x = 2( 4cos^3 y - 3cos y) = 2cos3y $ So ; the initial equation is equivalent to : $ cos 3y = cos \frac{y}{2} $ $ \iff 3y = \frac{y}{2} + k2 \pi $ or $ 3y = \frac{-y}{2} + m2 \pi$ ( $k ;m \in \mathbb{Z}$ ) $ \iff y = \frac{4k \pi}{5} $ or $ y = \frac{4m \pi}{7} $ ( $k ;m \in \mathbb{Z}$ ) With the condition $0 \le y \le \pi$ ; we can conclude that $ k;m \in \{ 0 ; 1 \} $ Thus ; the given equation has three distint root : $ x = 2cos \frac{4 \pi}{5} ; x = 2cos \frac{4 \pi}{7} ; x = 2$ ; Done