1 -43 When $x-y=1$, the value of $x^{4}-x y^{3}-x^{3} y-3 x^{2} y+3 x y^{2}+y^{4}$ is
(A) -1 .
(B) 0 .
(C) 1 .
(D) 2 .
(China Beijing Junior High School Mathematics Competition, 1991)
$$
\text { [Solution] Original expression } \begin{aligned}
& =x\left(x^{3}-y^{3}\right)-y\left(x^{3}-y^{3}\right)-3 x y(x-y) \\
& =\left(x^{3}-y^{3}\right)(x-y)-3 x y(x-y),
\end{aligned}
$$

Since $x-y=1$, the expression can be simplified to
$$
\begin{aligned}
\text { Original expression } & =\left(x^{3}-y^{3}\right)-3 x y=(x-y)\left(x^{2}+x y+y^{2}\right)-3 x y \\
& =x^{2}+x y+y^{2}-3 x y=(x-y)^{2}=1 .
\end{aligned}
$$

Therefore, the answer is $(C)$.
$$