\[y = \frac{x^{3}}{x - 4};\ \ \ \ y = x^{2} + 2x\]
\[\frac{x^{3}}{x - 4} = x^{2} + 2x^{\backslash x - 4};\ \ \ \ \ x \neq 4\]
\[x^{3} = x^{3} + 2x^{2} - 4x^{2} - 8x\]
\[x^{3} - x^{3} + 2x^{2} + 8x = 0\]
\[2x^{2} + 8x = 0\]
\[2x(x + 4) = 0\]
\[x = 0;\ \ \ \ \ x = - 4\]
\[y = 0;\ \ \ \ y = 16 - 8 = 8.\]
\[Ответ:графики\ пересекаются\ в\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ точках\ (0;0)\ и\ ( - 4;8).\]