Решите уравнение: x^3-3x^2+4x-12=0.

\[x^{3} - 3x^{2} + 4x - 12 = 0\]

\[x^{2}(x - 3) + 4(x - 3) = 0\]

\[(x - 3)\left( x^{2} + 4 \right) = 0\]

\[x - 3 = 0\ \ \ \ \ \ \ \ x^{2} + 4 = 0\]

\[x = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{2} = - 4\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ нет\ корней\]

\[Ответ:x = 3.\]