1. Given that $a$ and $b$ are distinct positive real numbers, and $a^{3}-b^{3}$ $=a^{2}-b^{2}$. Then the range of values for $a+b$ is ( ).
(A) $0<a+b<\frac{1}{3}$
(B) $1<a+b<\frac{4}{3}$
(C) $\frac{3}{4}<a+b<1$
(D) $\frac{1}{3}<a+b<2$
- 1. B.

From the given, it is easy to get $a^{2}+a b+b^{2}=a+b$, which means $\frac{a^{2}+a b+b^{2}}{a+b}=1$.
Since $(a+b)^{2}>a^{2}+a b+b^{2}$, therefore, $a+b>\frac{a^{2}+a b+b^{2}}{a+b}=1$.
Also, $(a-b)^{2}>0(a \neq b)$, then $3(a+b)^{2}<4\left(a^{2}+a b+b^{2}\right)$,
which means $a+b<\frac{4}{3}\left(\frac{a^{2}+a b+b^{2}}{a+b}\right)=\frac{4}{3}$.
Thus, $1<a+b<\frac{4}{3}$.