2.16. If $x \oplus y=\frac{4(x+10 y)}{x y+160}$. Answer: 0.26.

III. For each $a$, for which the equation $x^{3}-x^{2}-4 x-a=0$ has three distinct real roots, denote these roots, ordered in descending order $\left(x_{1}>x_{2}>x_{3}\right)$, by $x_{1}=x_{1}(a), x_{2}=x_{2}(a), x_{3}=x_{3}(a)$. Determine for which of these $a$ the expression $x_{1}^{2} x_{2}+x_{2}^{2} x_{3}+x_{3}^{2} x_{1}$ takes the maximum possible value. Round the answer to two decimal places if necessary. If there are multiple such $a$, indicate their sum in the answer.
Solution. Let's find for which $a$ there will be exactly three distinct solutions. For this, consider the expression $\tilde{f}(x)=x^{3}-x^{2}-4 x$. Three solutions $\tilde{f}(x)=a$ will exist if and only if the line $y=a$ intersects the graph of the function $y=\tilde{f}(x)$ at three points. Let's find the derivative $\tilde{f}(x)^{\prime}=3 x^{2}-2 x-4=$ $3\left(x-x_{+}\right)\left(x-x_{-}\right)$, where $x_{ \pm}=(1 \pm \sqrt{13}) / 3$. Since the sign of the derivative changes when passing through the roots $x_{ \pm}$, starting with a positive sign at $+\infty$, the local minimum $\tilde{f}_{\min }=\tilde{f}\left(x_{+}\right)=-\left(\frac{38+26 \sqrt{13}}{27}\right)=-4.879 \ldots$,
and the local maximum $\tilde{f}_{\max }=\tilde{f}\left(x_{-}\right)=\frac{-38+26 \sqrt{13}}{27}=2.064 \ldots$ Therefore, for all $a \in\left(\tilde{f}_{\min }, \tilde{f}_{\max }\right)$ and only for them, the equation $f(x)=0$ will have three distinct solutions. Let's introduce the notations

$$
\begin{aligned}
& u=x_{1}^{2} x_{2}+x_{2}^{2} x_{3}+x_{3}^{2} x_{1} \\
& v=x_{3}^{2} x_{2}+x_{2}^{2} x_{1}+x_{1}^{2} x_{3}
\end{aligned}
$$

From Vieta's theorem, it is known:

$$
\begin{array}{r}
\sigma_{1}=x_{1}+x_{2}+x_{3}=1 \\
\sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=-4 \\
\sigma_{3}=x_{1} x_{2} x_{3}=a
\end{array}
$$

It is easy to verify the validity of the equalities:

$$
\begin{array}{r}
u+v=\sigma_{1} \sigma_{2}-3 \sigma_{3} \\
u \cdot v=9 \sigma_{3}^{2}-6 \sigma_{1} \sigma_{2} \sigma_{3}+\sigma_{2}^{3}+\sigma_{3} \sigma_{1}^{3}
\end{array}
$$

From which

$$
\begin{array}{r}
u+v=-4-3 a \\
u \cdot v=9 a^{2}+25 a-64
\end{array}
$$

Therefore, the variables $u$ and $v$ are the solutions to the quadratic equation:

$$
u^{2}+(4+3 a) u+\left(9 a^{2}+25 a-64\right)=0
$$

Considering that $u-v=\left(x_{1}-x_{2}\right)\left(x_{1}-x_{3}\right)\left(x_{2}-x_{3}\right)>0$, we find the maximum value that $u$ can take

$$
u_{\max }=\frac{1}{2}\left(-4-3 a+\sqrt{-27 a^{2}-76 a+272}\right)
$$

To find the maximum value of the last expression for $a \in\left(\tilde{f}_{\min }, \tilde{f}_{\max }\right)$, we compute the derivative and set it to zero:

$$
-3+\frac{-27 a-38}{\sqrt{-27 a^{2}-76 a+272}}=0 \Longrightarrow-27 a-38=3 \sqrt{-27 a^{2}-76 a+272}
$$

The last equation is equivalent to the following: $243 a^{2}+684 a-251=0$ under the condition that $-27 a-38 \geqslant 0$. From this, $a=a^{*}=\frac{-38-13 \sqrt{13}}{27} \approx-3.1434 \ldots$ From the form of $a^{*}$, it follows that $a^{*} \in\left(\tilde{f}_{\min }, \tilde{f}_{\max }\right)$, and from the form of the derivative, it follows that for $a<a^{*}$ the derivative is positive in the domain of definition of the derivative, and for $a>a^{*}$ the derivative is negative in the domain of definition of the derivative. Therefore,

$$
u_{\max }=u\left(a^{*}\right) \approx 10.53
$$

Answer: $-3.14$.