Example 1 (Question from the 13th "Hope Cup" Invitational Competition) The real roots of the equations $x^{5}+x+1=0$ and $x+\sqrt[5]{x}+1=0$ are $\alpha, \beta$ respectively, then $\alpha+\beta$ equals ( ).
A. -1
B. $-\frac{1}{2}$
C. $\frac{1}{2}$
D. 1
Solution: Choose A. Reason: Consider the function $f(x)=x^{5}+x+1$, which is an increasing function on $\mathbf{R}$,
when $x^{5}+x+1=0$,
we get $x=\sqrt[5]{-x-1}$, substituting $-x-1$ for $x$ yields $-x-1=\sqrt[5]{x}$.
From (2), we have $(-x-1)^{5}=x$, which is also $(-x-1)^{5}+(-x-1)+1=0$.
Let $\alpha$ be a root of equation (1), i.e., $f(\alpha)=0$. Let $\beta$ be a root of equation (2), i.e., $-1-\beta=\sqrt[5]{\beta}$, thus from (3) we know that $f(-1-\beta)=0$.
Therefore, we have $f(\alpha)=f(-1-\beta)$, which means $\alpha+\beta=-1$.