Solve the equation $\sqrt{x-3}+\sqrt{x^{2}-4x+3}=\sqrt{(x-2)^{3}}$.
Solution. With the substitution $A=x-2$, the equation becomes

$$
\sqrt{A-1}+\sqrt{A^{2}-1}=\sqrt{A^{3}}
$$

The equation is only meaningful for $A \geq 1$ (i.e., $x \geq 3$). Squaring both sides, we get:

$$
A-1+2 \sqrt{A-1} \sqrt{A^{2}-1}+A^{2}-1=A^{3}
$$

Rearranging terms, we have:

$$
2 \sqrt{A-1} \sqrt{A^{2}-1}=A^{3}-A^{2}-A+2
$$

Now, let's introduce a new variable, $B:=\sqrt{(A-1)\left(A^{2}-1\right)}$. Multiplying the two polynomials under the square root, $(A-1)\left(A^{2}-1\right) = A^{3}-A^{2}-A+1$; we see that the right-hand side of the equation can also be rewritten as:

$$
2 B=B^{2}+1
$$

From this, we immediately get $(B-1)^{2}=0, B=1$. This implies:

$$
1=A^{3}-A^{2}-A+1, \quad A^{3}-A^{2}-A=0, \quad A\left(A^{2}-A-1\right)=0
$$

Since $A \geq 1$, we only need to examine the roots of $A^{2}-A-1=0$. Among these, $\frac{1-\sqrt{5}}{2}$ is less than 1, so only $A=\frac{1+\sqrt{5}}{2}$ can be a solution. From this,

$$
x=A+2=\frac{1+\sqrt{5}}{2}+2=\frac{5+\sqrt{5}}{2}
$$

which is indeed a root of the original equation.