<s>[INST] How to solve the equation $x^2+2=4\sqrt{x^3+1}$? From the Leningrad Mathematical Olympiad, 1975:

Solve $x^2+2=4\sqrt{x^3+1}$.

In answer sheet only written $x=4+2\sqrt{3}\pm \sqrt{34+20\sqrt{3}}$. 
How to solve this?
 [/INST] Let $u = x^2-x+1$, $v = x+1$  and $\lambda = \sqrt{\frac{u}{v}}$, we have
$$\begin{align}u + v & = (x^2-x+1) + (x + 1) = x^2 + 2 \\ &= 4\sqrt{x^3+1} = 4\sqrt{(x+1)(x^2-x+1)}\\ &= 4\sqrt{uv}\end{align}$$
This leads to
$$\lambda^2 + 1 = 4\lambda \implies \lambda = 2 \color{red}{\pm} \sqrt{3}$$
As a result,
$$\begin{align} &x^2 - x + 1 = \lambda^2 (x+1) = (7\color{red}{\pm} 4\sqrt{3})(x+1)\\
\iff & x^2 - (8 \color{red}{\pm} 4\sqrt{3})x  = (6\color{red}{\pm} 4\sqrt{3})\\
\iff & (x - (4 \color{red}{\pm} 2\sqrt{3}))^2  = (4 \color{red}{\pm} 2\sqrt{3})^2 + (6 \color{red}{\pm} 4\sqrt{3})
= 34 \color{red}{\pm} 20\sqrt{3}\\
\implies & x = 4 \color{red}{\pm} 2\sqrt{3} \color{blue}{\pm} \sqrt{34 \color{red}{\pm} 20\sqrt{3}}
\end{align}$$
Notice $34-20\sqrt{3} \approx -0.651 < 0$. For real $x$, 
the red $\color{red}{\pm}$ sign need to be $+$. This leaves us with
$$x = 4 + 2\sqrt{3} \color{blue}{\pm} \sqrt{34 + 20\sqrt{3}}$$
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