Решите уравнение: x^3-3x^2-3x+1=0.

\[x^{3} - 3x^{2} - 3x + 1 = 0\]

\[\left( x^{3} + 1 \right) - \left( 3x^{2} + 3x \right) = 0\]

\[(x + 1)\left( x^{2} - x + 1 \right) - 3x(x + 1) = 0\]

\[(x + 1)\left( x^{2} - x + 1 - 3x \right) = 0\]

\[1)\ x + 1 = 0\]

\[x = - 1.\]

\[2)\ x^{2} - 4x + 1 = 0\]

\[D_{1} = 4 - 1 = 3\]

\[x_{1,2} = 2 \pm \sqrt{3}.\]

\[Ответ:x = - 1;x = 2 \pm \sqrt{3}.\]