Problem 870

Answer: $(-\infty,-8) \cup(-20 / 3,61 / 8)$.

Solution. Note that $x=0$ is not a solution to the original equation. Therefore, it is equivalent to the equation $a=\frac{-x^{3}-13 x+6}{x^{2}}(*)$. Let's denote the right side by $f(x)$.

Note that $f(x) \xrightarrow{\rightarrow}+\infty$ 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 $f^{\prime}(x)=-\frac{x^{3}-13 x+12}{x^{3}}=-\frac{(x-3)(x-1)(x+4)}{x^{3}}$.

Thus, the function $f(x)$ on the interval $(-\infty,-4]$ decreases from $+\infty$ to $f(-4)=61 / 8$; on the interval $[-4,0)$ - increases from $61 / 8$ to $+\infty$; on the interval $(0,1]-$ decreases from $+\infty$ to $f(1)=-8$; on the interval $[1,3]$ - increases from -8 to $f(3)=-20 / 3$; finally, on the interval $[3,+\infty)-$ decreases from $-20 / 3$ to $-\infty$.

Therefore, each value from the interval ( $-\infty ;-8$ ) is taken by the function $f(x)$ exactly once; -8 - twice; from the interval ( $-8 ;-20 / 3$ ) - three times; $-20 / 3$ - twice; from the interval $(-20 / 3 ; 61 / 8)$ - once; $61 / 8$ - twice; from the interval ( $61 / 8 ;+\infty)$ - three times. (For example, the value -8 will be taken by the function once at the point 1, and the second time - on the interval $(3,+\infty)$ ).

Thus, the equation $\left({ }^{*}\right)$, and with it the original equation, has a unique solution when $a \in(-\infty,-8) \cup(-20 / 3,61 / 8)$.