2021/049) Solve the equation $x^3-3x= \sqrt{(x+2)}$

 As RHS is not negative so we have $x^2>=3$

Trying the lowest integer $>\sqrt{3}$ that is 2 we get this satisfies the equation.

Now squaring both sides we get $x^2(x^2-3)^2 = x + 2$

at $x >2$ LHS grows faster than RHS so x = 2 is the only solution