\section*{Solution}

Put $y=2-h$, where $p(h)=1$, then $(2-y)^{3}-3(2-y)^{2}+5(2-y)-1=0$, so $8-12 y+6 y^{2}-y^{3}-12+12 y-$ $3 y^{2}+10-5 y-1=0$, or $y^{3}-3 y^{2}+5 y=5$, or $p(y)=5$. So if $h$ is a root of $p(h)=1$, then there is a root $k$ of $p(k)=5$ such that $h+k=2$. To complete the proof we have to show that $p(x)=5$ has only one real root.

But $x^{3}-3 x^{2}+5 x=(x-1)^{3}+2(x-1)+3$ which is a strictly increasing function of $x-1$ and hence of $x$. So $p(x)=k$ has only one real root.