256. How can a cubic equation be quickly solved if one of its roots is known? Provide numerical examples.
256. If given a cubic equation $x^{3}+a x^{2}+b x+c=0$, and if $\alpha$ is a root of this equation, then, as is known, the left side of the equation is divisible by $x-\alpha$. If one of the roots $\alpha$ is known, then by dividing the left side of the equation by $x-\alpha$, we obtain a quadratic equation, from which we can easily determine the other two roots. One, for example, of the roots of the equation $x^{3}-6 x^{2}+11 x-6=0$ is $x_{1}=1$; dividing the left side of the equation by $x-1$, we get $x^{2}-5 x+6$, from which: $x_{2}=2, x_{3}=3$.