<s>[INST] Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$. 
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$

Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x - 3)^2 \implies a + b = \dfrac{a^2 - b^2}{a + b} = \dfrac{2x - 1}{(x - 3)^2}$.
$\left\{
\begin{align}
a = \dfrac{(a + b) + (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 106x + 80}{2(x - 3)^2}\\
b = \dfrac{(a + b) - (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 110x + 82}{2(x - 3)^2}
\end{align}
\right.$
 [/INST] $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9 \implies (\sqrt{x + 2} - 2) - (\sqrt{3 - x} - 1) = x^2 - 6x + 8$$
$$ \implies (x - 2)\left(\frac{1}{\sqrt{x + 2} + 2} + \frac{1}{\sqrt{3 - x} + 1}\right) = (x - 2)(x - 4)$$
$$\implies \left[ \begin{align*} x - 2 &= 0\\ \frac{1}{\sqrt{x + 2} + 2} + \frac{1}{\sqrt{3 - x} + 1} &= x - 4 \end{align*} \right.$$
For the second equality, we have that $-2 \le x \le 3$ and $x \ge 4$ as the needed condition for the left and right side of the equality. So there're not any solutions for the second equality.
That means $x - 2 = 0 \implies x = 2$.
</s>