Problem 8. For what values of the parameter $a$ does the equation $x^{3}+6 x^{2}+a x+8=0$ have no more than two solutions?
Answer: $[-15 ;+\infty)$.

Solution. Note that $x=0$ is not a solution to the original equation. Therefore, it is equivalent to the equation $a=\frac{-x^{3}-6 x^{2}-8}{x}(*)$. Let's denote the right side by $f(x)$.

Notice that $f(x) \xrightarrow{x}-\infty$ as $x \rightarrow-\infty$ and $f(x) \rightarrow-\infty$ as $x \rightarrow+\infty$. Also, $f(x)$ has a vertical asymptote at $x=0$.

The derivative of the function $f(x)$ is $f^{\prime}(x)=\frac{8}{x^{2}}-2 x-6=\frac{8-2 x-6 x^{2}}{x^{2}}=-\frac{2(x-1)(x+2)^{2}}{x^{2}}$.

Thus, the function $f(x)$ on the interval $(-\infty,-2]$ increases from $-\infty$ to $f(-2)=12$; on the interval $[-2,0)$ - continues to increase from 12 to $+\infty$; on the interval $(0,1]-$ increases from $-\infty$ to $f(1)=$ $-15 ;$ finally, on the interval $[1,+\infty)$ - decreases from -15 to $-\infty$.

Therefore, each value from the interval ( $-\infty ;-15$ ) the function $f(x)$ takes exactly three times; -15 - two times; from the interval $(-15 ;+\infty)$ - once. (For example, the value -15 the function will take once at the point 1, and the second time - on the interval $(-\infty,-2)$ ).

Hence, the equation $\left(*^{*}\right.$ ), and with it the original equation, has no more than two solutions when $a \in[-15 ;+\infty)$.