Consider the cubic equation x^3-6x^2+9x-k=0, where k is a real constant.

Let $h(x)=x^{3}-6x^{2}+9x$. Then $h'(x)=3(x-1)(x-3)$, so $h$ has a local maximum $h(1)=4$ and a local minimum $h(3)=0$. The equation $h(x)=k$ has three distinct real roots iff $0<k<4$. (A) False: $k=5>4$ gives one real root. (B) True: $k=4$ touches the local maximum, so $x^{3}-6x^{2}+9x-4=(x-1)^{2}(x-4)$ with distinct real roots $1$ and $4$. (C) False: $k=-1<0$ gives only one real root (the other two are complex). (D) False: for any $k$, by Vieta $\sum x_i=6$ and $\sum x_ix_j=9$, so $\sum x_i^{2}=36-18=18\ne20$. Hence only (B) is correct.