Example 10. Let real numbers $x, y$ satisfy the equation $x^{3}+y^{3}=a^{3}(a>0)$. Find the range of values for $x+y$:
(1992, Taiyuan City Junior High School Mathematics Competition)
Let $x+y=t$. Then
$$
\begin{aligned}
x^{3}+y^{3} & =(x+y)^{3}-3(x+y) x y \\
& =t^{3}-3 t x y=a^{3} . \\
\therefore \quad x y & =\frac{t^{3}-a^{3}}{3 t} .
\end{aligned}
$$

From this, we know that $x, y$ are the two real roots of the equation about $z$
$$
z^{2}-t z+\frac{t^{3}-a^{3}}{3 t}=0
$$

By the discriminant theorem, we get
$$
\Delta=t^{2}-4 \cdot \frac{t^{3}-a^{3}}{3 t} \geqslant 0 .
$$

Solving this, we get $0<t \leqslant \sqrt[3]{4} a$.
That is, $0<x+y \leqslant \sqrt[3]{4} a$.