Решите уравнение: x^3-43x+42=0.

\[x^{3} - 43x + 42 = 0\]

\[x^{3} - x - 42x + 42 = 0\]

\[x\left( x^{2} - 1 \right) - 42(x - 1) = 0\]

\[x(x - 1)(x + 1) - 42(x - 1) = 0\]

\[(x - 1)\left( x(x + 1) - 42 \right) = 0\]

\[(x - 1)\left( x^{2} + x - 42 \right) = 0\]

\[x = 1;\]

\[x^{2} + x - 42 = 0\]

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 42\]

\[x_{1} = - 7;\ \ \ x_{2} = 6.\]

\[Ответ:x = - 7;\ \ x = 1;\ \ x = 6.\]