\[\boxed{\mathbf{893}\mathbf{.}}\]
\[f(x) = x^{3} - px\ \ x_{0} = 1\]
\[M(2\ 3):\]
\[f^{'}(x) = \left( x^{3} \right)^{'} - p \bullet (x)^{'} = 3x^{2} - p\]
\[f^{'}(1) = 3 \bullet 1^{2} - p = 3 - p\]
\[f(1) = 1^{3} - p \bullet 1 = 1 - p\]
\[y = 1 - p + (3 - p)(x - 1) =\]
\[= 1 - p + 3x - 3 - px + p =\]
\[= 3x - px - 2.\]
\[Проходит\ через\ точку\ M:\]
\[3 = 3 \bullet 2 - p \bullet 2 - 2\]
\[6 - 2p = 5\]
\[1 = 2p\]
\[p = \frac{1}{2}.\]
\[Ответ:\ \ p = 0,5.\]