3. Solve the equation

$$
2(x-6)=\frac{x^{2}}{(1+\sqrt{x+1})^{2}}
$$
Solution 1. Rewrite the equation as $2(x-6)(1+\sqrt{x+1})^{2}=x^{2}$ and make the substitution $\sqrt{x+1}=y$. Then $x=y^{2}-1$ and $y \geq 0$. After the substitution, we get $2\left(y^{2}-7\right)(y+1)^{2}=$ $\left(y^{2}-1\right)^{2}$. Factor the right side as $(y+1)^{2}(y-1)^{2}$, move it to the left, and factor out $(y+1)^{2}$. We get $(y+1)^{2}\left(2 y^{2}-14-(y-1)^{2}\right)=(y+1)^{2}\left(y^{2}+\right.$ $2 y-15)=(y+1)^{2}(y-3)(y+5)=0$. This equation has three solutions $y_{1}=3, y_{2}=$ $-1, y_{3}=-5$. The last two do not satisfy the condition $y \geq 0$, so they are extraneous. If $y=3$, then $x=8$. Clearly, this solution is valid.

Solution 2. Transform the equation to the form

$$
2(x-6)(x+2+2 \sqrt{x+1})-x^{2}=0
$$

and denote the expression on the left side by $f(x)$. We have

$$
f(x)=2 x^{2}-8 x-24+4(x-6) \sqrt{x+1}-x^{2}=x^{2}-8 x-24+4(x-6) \sqrt{x+1}
$$

We need to solve the equation $f(x)=0$. It is clear that all roots of this equation satisfy the condition $x>6$ (otherwise, the left side of the original equation is non-positive, while the right side is positive). Find the derivative $f^{\prime}(x)$ :

$$
f^{\prime}(x)=(2 x-8)+4 \sqrt{x+1}+\frac{2(x-6)}{\sqrt{x+1}}
$$

It is clear that for $x>6$ each term in the last sum is non-negative. Therefore, the function $f(x)$ is increasing on the interval $(6,+\infty)$. Hence, the equation $f(x)=$ 0 can have no more than one solution. One solution $x=8$ can be easily found by trial.