2.104 If $p, q$ and $r$ are the distinct roots of $x^{3}-x^{2}+x-2=0$, then $p^{3}+q^{3}+r^{3}$ equals
(A) -1 .
(B) 1 .
(C) 3 .
(D) 5 .
(E) None of these.
(26th American High School Mathematics Examination, 1975)
［Solution］ Substituting the identity
$$
p^{2}+q^{2}+r^{2}=(p+q+r)^{2}-2(p q+q r+r p) \text {, }
$$

into another identity
$$
p^{3}+q^{3}+r^{3}=(p+q+r)\left(p^{2}+q^{2}+r^{2}-p q-q r-r p\right)+3 p q r \text {, }
$$

and using Vieta's formulas, we have
$$
\begin{aligned}
p^{3}+q^{3}+r^{3} & =(p+q+r)\left[(p+q+r)^{2}-3(p q+q r+r p)\right]+3 p q r \\
& =1 \cdot\left(1^{2}-3 \cdot 1\right)+3 \cdot 2=4 .
\end{aligned}
$$

Therefore, the answer is $(E)$.