$$\begin{array}{l}
{\text{Известно}}{\text{, что}} \hfill \\
\left\{ \begin{gathered}
{c^3} - 3{c^2} + 5c - 17 = 0 \hfill \\
{d^3} - 3{d^2} + 5d + 11 = 0 \hfill \\
\end{gathered} \right.. \hfill \\
{\text{Найдите }}c + d. \hfill \\
\end{array} $$
\[\begin{array}{l}
\left\{ \begin{gathered}
{c^3} - 3{c^2} + 5c - 3 = 14 \hfill \\
{d^3} - 3{d^2} + 5d - 3 = - 14 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
\left( {c - 1} \right)\left( {{{\left( {c - 1} \right)}^2} + 2} \right) = 14 \hfill \\
\left( {1 - d} \right)\left( {{{\left( {1 - d} \right)}^2} + 2} \right) = 14 \hfill \\
\end{gathered} \right. \Rightarrow \hfill \\
\left( {c - 1} \right)\left( {{{\left( {c - 1} \right)}^2} + 2} \right) = \left( {1 - d} \right)\left( {{{\left( {1 - d} \right)}^2} + 2} \right){\text{, т}}{\text{.е}}{\text{.}} \hfill \\
{\text{получили уравнение вида}} \hfill \\
f\left( {c - 1} \right) = f\left( {1 - d} \right){\text{, где }}f\left( x \right) = x\left( {{x^2} + 2} \right){\text{ - монотонная функция}}{\text{.}} \hfill \\
{\text{Но тогда}} \hfill \\
c - 1 = 1 - d \Leftrightarrow c + d = 2. \hfill \\
\end{array} \]
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