7-70 If $x, y$ are real numbers, and $x^{3}+y^{3}=2$, then the range of $x+y$ is
(A) $(-\infty, 2)$.
(B) $[-\sqrt[3]{2}, \sqrt[3]{2}]$.
(C) $(0,2]$.
(D) $(2,+\infty)$.
(China Beijing High School Mathematics Competition, 1993)
[Solution 1] Consider two cases:
(1) If $x \geqslant y \geqslant 0, x^{3}, y^{3}$ are equivalent to the volumes of cubes with edge lengths $x, y$ respectively. For a cube with a total volume of 2, consider two extreme scenarios:
(1) $x=\sqrt[3]{2}, y=0, \quad$ in this case $x+y=\sqrt[3]{2}$;
(2) $x=y=1$, in this case $x+y=2$.
Thus, $x+y$ varies between $\sqrt[3]{2}$ and 2.
(2) If $x>0, y>0$.
In summary, $x+y \in(0,2]$.
Therefore, the answer is $(C)$.
[Solution 2] Given $x^{3}+y^{3}=2$,

which is $(x+y)\left(x^{2}-x y+y^{2}\right)=2$,
so $x+y=\frac{2}{x^{2}-x y+y^{2}}$.
If $x>0, y>0$, then when $x=y=1$, $(x+y)_{\text {max }}=2$.
If $x>0, y>0$, then $t=x+y>0$ always holds $(t \in R)$,
$\therefore t(t-2) \leqslant 0$, i.e., $0 \leqslant t \leqslant 2$.
When $t=0$, $x=-y$, then $x^{3}+y^{3}=0$ which contradicts the given condition.
Thus, $0 < t \leqslant 2$.
Since $x>0, y>0, \quad \therefore \quad x+y>0$, thus options $(A),(B)$ can be ruled out.
If $x=0, y=\sqrt[3]{2}$, then $x+y=\sqrt[3]{2} \in(0,2]$ which rules out $(D)$. Therefore, the answer is $(C)$.