$1 \cdot 41$ If the sum of two numbers is 1, and the product is 1, then the sum of the cubes of these two numbers is
(A) 2.
(B) $-2-\frac{3 \sqrt{3} i}{4}$.
(C) 0.
(D) $-\frac{3 \sqrt{3}}{4} i$.
(E) -2.
(17th American High School Mathematics Examination, 1966)
[Solution] Let the two numbers be $x, y$. Given $x+y=x y=1$. (Note that $x, y$ are not real numbers)

Also, $1=(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)=x^{3}+y^{3}+3$, then $x^{3}+y^{3}=-2$.
Therefore, the answer is $(E)$.