6. The number of all distinct real solutions of the equation $2 x^{3}+6 x^{2}+12 x+8+x^{3} \sqrt{x^{2}+1}$ $+(x+2)^{3} \sqrt{(x+2)^{2}+1}=0$ is $\qquad$.
6.1.

Rewrite the equation as
$$
\begin{array}{l}
x^{3}+x^{3} \sqrt{x^{2}+1}+(x+2)^{3}+ \\
(x+2)^{3} \sqrt{(x+2)^{2}+1}=0 . \\
\text { Let } f(x)=x^{3}+x^{3} \sqrt{x^{2}+1} \\
=x^{3}\left(1+\sqrt{x^{2}+1}\right) .
\end{array}
$$

Then equation (1) can be rewritten as
$$
f(x+2)=-f(x) \text {. }
$$

Clearly, $f(-x)=-f(x)$.
From equations (2) and (3), we get
$$
f(x+2)=f(-x) \text {. }
$$

Clearly, $g(x)=x^{3}$ is an increasing function, and $h(x)=1+\sqrt{x^{2}+1}$ is an increasing function for $x \geqslant 0$ and a decreasing function for $x \leqslant 0$, and $h(x) \geqslant 2(x \in \mathbf{R})$. Therefore, when $x \geqslant 0$, $f(x)=g(x) h(x)$ is an increasing function; when $x \leqslant 0$, $f(x)=g(x) h(x)$ is also an increasing function.

If $x_{2}<0<x_{1}$, then $f\left(x_{2}\right)<f(0)<f\left(x_{1}\right)$. Therefore, $f(x)$ is an increasing function on $\mathbf{R}$.
Thus, from equation (4) we get $x+2=-x$, i.e., $x=-1$.
Clearly, $x=-1$ satisfies equation (1). Hence, the given equation has a unique real root $x=-1$.