\[\boxed{\text{244.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[Пусть\ x - это\ искомое\ число.\ \]
\[Тогда\ 13 \cdot \left( x + \frac{1}{x} \right) = x^{3} + \frac{1}{x^{3}}.\]
\[Пусть\ \ x + \frac{1}{x} = t,\ \ t > 0,\]
\[\ \ тогда:\]
\[x^{3} + \frac{1}{x^{3}} = \left( x + \frac{1}{x} \right)\left( x^{2} - 1 + \frac{1}{x^{2}} \right)\]
\[x^{2} + \frac{1}{x^{2}} = \left( x + \frac{1}{x} \right)^{2} - 2 =\]
\[= t^{2} - 2\]
\[x^{3} + \frac{1}{x^{3}} = t \cdot \left( t^{2} - 2 - 1 \right) =\]
\[= t \cdot \left( t^{2} - 3 \right).\]
\[\Longrightarrow 13t = t \cdot \left( t^{2} - 3 \right)\]
\[13t = t^{3} - 3t\]
\[t³ - 16t = 0\]
\[t\left( t^{2} - 16 \right) = 0\]
\[t(t - 4)(t + 4) = 0\]
\[t_{1} = 0,\ \ t_{2,3} = \pm 4.\]
\[Так\ как\ t > 0,\ то\ \ \]
\[\ x + \frac{1}{x} = 4\]
\[x^{2} - 4x + 1 = 0,\]
\[x_{1,2} = 2 \pm \sqrt{3}.\]
\[Ответ:x = 2 \pm \sqrt{3}.\]