9.1 Solve the equation $x(x-1)(x-2)+(100-x)(99-x)(98-x)=0$.
Answer: no solutions. Note that the given equation is quadratic (the coefficients of $x^{3}$ cancel out). The vertex of this quadratic trinomial is at the point with abscissa $x_{0}=\frac{100}{2}=50$; this can be verified either by direct calculation or in the following way: if we introduce the substitution $t=x-$ 50, then the quadratic trinomial can be written as $f(t)=(50+t)(49+t)(48+t)+$ $(50-\mathrm{t})(49-\mathrm{t})(48-\mathrm{t})$, and it is easy to see that the coefficients of the odd powers of $t$ cancel out, and the value at $t=0$ (the ordinate of the vertex) is positive, and the leading coefficient (of $t^{2}$) is also positive. Therefore, the equation has no roots.