Example 3 Given $p^{3}+q^{3}=2$, where $p, q$ are real numbers. Then the maximum value of $p+q$ is $\qquad$
(1987, Jiangsu Province Junior High School Mathematics Competition)
Solution: Let $s=p+q$.
From $p^{3}+q^{3}=2$ we get
$(p+q)\left(p^{2}+q^{2}-p q\right)=2$,
$(p+q)^{2}-3 p q=\frac{2}{s}$,
Thus, $p q=\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)$.
From (1) and (2), $p$ and $q$ are the two real roots of the equation
$$
x^{2}-s x+\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)=0
$$

We know $\Delta=s^{2}+\frac{4}{s}\left(s^{2}-\frac{2}{s}\right) \geqslant 0$.
Simplifying, we get $s^{2} \leqslant 8$.
Therefore, $s \leqslant 2$.
Taking $p=q=1$, the maximum value of $p+q$ is 2.
4 Using the definition of absolute value
That is, $|x|= \pm x,|x|^{2}=x^{2}$.