Решите уравнение: (x^2-1)(x^2+4x+3)=0.

\[\left( x^{2} - 1 \right)\left( x^{2} + 4x + 3 \right) = 0\]

\[x² - 1 = 0\]

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

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

\[D = 4^{2} - 4 \cdot 1 \cdot 3 = 16 - 12 = 4\]

\[x_{1} = \frac{- 4 + \sqrt{4}}{2} = \frac{- 4 + 2}{2} = \frac{- 2}{2} =\]

\[= - 1\]

\[x_{2} = \frac{- 4 - \sqrt{4}}{2} = \frac{- 4 - 2}{2} = \frac{- 6}{2} =\]

\[= - 3\]

\[Ответ:\ - 1;\ - 3;\ \ 1.\]