Решите уравнение: (x^2-16)(x^2+2x-8)=0.

\[\left( x^{2} - 16 \right)\left( x^{2} + 2x - 8 \right) = 0\]

\[(x - 4)(x + 4)\left( x^{2} + 2x - 8 \right) = 0\]

\[x^{2} + 2x - 8 = 0\]

\[D = 2^{2} - 4 \cdot 1 \cdot ( - 8) = 4 + 32 =\]

\[= 36\]

\[x_{1} = \frac{- 2 + \sqrt{36}}{2} = \frac{- 2 + 6}{2} = \frac{4}{2} =\]

\[= 2\]

\[x_{2} = \frac{- 2 - \sqrt{36}}{2} = \frac{- 2 - 6}{2} =\]

\[= \frac{- 8}{2} = - 4\]

\[Ответ:4;\ - 4;\ 2.\]