9.079. $\frac{x^{3}-x^{2}+x-1}{x+8} \leq 0$.

9.079. $\frac{x^{3}-x^{2}+x-1}{x+8} \leq 0$.
## Solution.

Domain of definition: $x \neq-8$.

Rewrite the inequality as $\frac{x^{2}(x-1)+(x-1)}{x+8} \leq 0, \frac{(x-1)\left(x^{2}+1\right)}{x+8} \leq 0$. Since $x^{2}+1>0$ for $x \in R$, the obtained inequality is equivalent to the system of inequalities $\left\{\begin{array}{l}(x-1)(x+8) \leq 0, \\ x \neq-8 .\end{array}\right.$

Using the interval method, we get $x \in(-8 ; 1]$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-545.jpg?height=168&width=502&top_left_y=1212&top_left_x=365)

Answer: $x \in(-8 ; 1]$