6.182. $|x|^{3}+|x-1|^{3}=9$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.

6.182. $|x|^{3}+|x-1|^{3}=9$.
## Solution.

As in № 6.179, we have three cases

1) $\left\{\begin{array}{l}x<0, \\ -x^{3}-(x-1)^{3}=9\end{array} \Leftrightarrow\left\{\begin{array}{l}x<0, \\ 2 x^{3}-3 x^{2}+3 x+8=0 .\end{array}\right.\right.$

By checking, we find that $x_{1}=-1$, since $-2-3-3+8=0$. We divide the left side of the equation by $x+1: \frac{2 x^{3}-3 x^{2}+3 x+8}{x+1}=2 x^{2}-5 x+8$.

The equation can be written as $(x+1)\left(2 x^{2}-5 x+8\right)=0$. In the equation $2 x^{2}-5 x+8=0$, $D<0$ and there are no roots.
2) $\left\{\begin{array}{l}0 \leq x<1, \\ x^{3}-(x-1)^{3}=9\end{array} \Leftrightarrow\left\{\begin{array}{l}0 \leq x<1, \\ 3 x^{2}-3 x-8=0 .\end{array}\right.\right.$

$x_{1,2}=\frac{3 \pm \sqrt{105}}{6}$. There are no solutions on the second interval because neither $x_{1}=\frac{3-\sqrt{105}}{6}$ nor $x_{2}=\frac{3+\sqrt{105}}{6}$ belong to $[0 ; 1)$;
3) $\left\{\begin{array}{l}x \geq 1, \\ x^{3}+(x-1)^{3}=9\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 1, \\ 2 x^{3}-3 x^{2}+3 x-10=0 .\end{array}\right.\right.$

By checking, we find that $x_{1}=2$, since $16-12+6-10=0$. We divide the left side of the equation by $x-2: \frac{2 x^{3}-3 x^{2}+3 x-10}{x-2}=2 x^{2}+x+5$.

The equation can be represented as $(x-2)\left(2 x^{2}+x+5\right)=0$, from which the root $x_{1}=2$ of the given equation is found by trial. In the equation $2 x^{2}+x+5=0$, $D<0$ and there are no roots.

Answer: $x_{1}=-1, x_{2}=2$.

Solve the systems of equations ( $6.183-6.243)$ :