Решите уравнение: 3x^3-7x^2-7x+3=0.

\[3x^{3} - 7x^{2} - 7x + 3 = 0\]

\[3 \bullet \left( x^{3} + 1 \right) - 7x(x + 1) = 0\]

\[(x + 1)\left( 3x^{2} - 3x + 3 - 7x \right) = 0\]

\[(x + 1)\left( 3x^{2} - 10x + 3 \right) = 0\]

\[x + 1 = 0 \Longrightarrow x = - 1.\]

\[3x^{2} - 10x + 3 = 0\]

\[D = ( - {10)}^{2} - 4 \cdot 3 \cdot 3 =\]

\[= 100 - 36 = 64\]

\[x_{1} = \frac{10 + \sqrt{64}}{2 \cdot 3} = \frac{10 + 8}{6} = \frac{18}{6} =\]

\[= 3\]

\[x_{2} = \frac{10 - \sqrt{64}}{2 \cdot 3} = \frac{10 - 8}{6} = \frac{2}{6} =\]

\[= \frac{1}{3}\]

\[Ответ:\ x = - 1;\ x = 3;\ \ x = \frac{1}{3}.\]