6.176. $(\sqrt{x+1}+\sqrt{x})^{3}+(\sqrt{x+1}+\sqrt{x})^{2}=2$.

6.176. $(\sqrt{x+1}+\sqrt{x})^{3}+(\sqrt{x+1}+\sqrt{x})^{2}=2$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x+1 \geq 0, \\ x \geq 0\end{array} \Leftrightarrow x \geq 0\right.$.

Let $\sqrt{x+1}+\sqrt{x}=y$, where $y \geq 0$. The equation in terms of $y$ becomes

$$
\begin{aligned}
& y^{3}+y^{2}-2=0 \Leftrightarrow y^{3}-1+y^{2}-1=0 \Leftrightarrow \\
& \Leftrightarrow(y-1)\left(y^{2}+y+1\right)+(y+1)(y-1)=0 \Leftrightarrow \\
& \Leftrightarrow(y-1)\left(y^{2}+y+1+y+1\right)=0,(y-1)\left(y^{2}+2 y+2\right)=0
\end{aligned}
$$

The obtained equation is equivalent to two equations: $y-1=0$ and $y^{2}+2 y+2=0$, solving which, we find $y_{1}=1$; the second equation has no solutions, since $D<0$.

Then $\sqrt{x+1}+\sqrt{x}=1 \Leftrightarrow x+1+2 \sqrt{(x+1) x}+x=1, \sqrt{(x+1) x}=-x$, where $x \leq 0$. From the domain of definition, we get $\left\{\begin{array}{l}x \geq 0 \\ x \leq 0\end{array} \Leftrightarrow x=0\right.$.

Answer: $x=0$.