$1 \cdot 38$ If $\left(r+\frac{1}{r}\right)^{2}=3$, then $r^{3}+\frac{1}{r^{3}}$ equals
(A) 1 .
(B) 2 .
(C) 0 .
(D) 3 .
(E) 6 .
(3rd American High School Mathematics Examination, 1952)
[Solution] From the assumption $\left(r+\frac{1}{r}\right)^{2}=3$, we know $r^{2}-1+\frac{1}{r^{2}}=0$, thus $r^{3}+\frac{1}{r^{3}}=\left(r+\frac{1}{r}\right)\left(r^{2}-1+\frac{1}{r^{2}}\right)=0$.

Therefore, the answer is $(C)$.