10・43 If the distinct non-zero numbers $x(y-z), y(z-x), \quad z(x-y)$ form a geometric sequence with common ratio $r$, then $r$ satisfies the equation
(A) $r^{2}+r+1=0$.
(B) $r^{2}-r+1=0$.
(C) $r^{4}+r^{2}-1=0$.
(D) $(r+1)^{4}+r=0$.
(E) $(r-1)^{4}+r=0$.
(29th American High School Mathematics Examination, 1978)
[Solution] Let $a=x(y-z)$ and note the following identity
$$
x(y-z)+y(z-x)+z(x-y)=0 \text {. }
$$

Therefore,
$$
a+a r+a r^{2}=0 \text {, }
$$

which means $\square$
$$
1+r+r^{2}=0 \text {. }
$$

Thus, the answer is $(A)$.