Example 9 Find the real roots of the following equations:
(1) $\sqrt[3]{x+1}+\sqrt[3]{2 x+3}+3 x+4=0$;
(2) $126 x^{3}+225 x^{2}+141 x+30=0$.
(1) The original equation can be transformed into $\sqrt[3]{x+1}+x+1+\sqrt[3]{2 x+3}+2 x+3=0$.
Let $x+1=t$, then the equation can be transformed into $\sqrt[3]{t}+t+\sqrt[3]{2 t+1}+2 t+1=0$.
Consider the function $f(t)=\sqrt[3]{t}+t$, it is easy to know that $f(t)$ is an odd function, and is monotonically increasing on $\mathbf{R}$. At this point, the original equation can be transformed into $f(t)+f(2 t+1)=0$, that is, $f(2 t+1)=-f(t)=f(-t)$.
Thus, by the conclusion $3(1)$ of this chapter, we have $2 t+1=-t$,
which gives $t=-\frac{1}{3}$, so $x=-\frac{4}{3}$ is the solution.
(2) The original equation can be transformed into $(5 x+3)^{3}+x^{3}+(5 x+3)+x=0$.
Consider the function $f(x)=x^{3}+x$, it is easy to know that $f(x)$ is an odd function, and is monotonically increasing on $\mathbf{R}$. At this point, the original equation can be transformed into $f(5 x+3)+f(x)=0$, that is, $f(5 x+3)=-f(x)=f(-x)$.
Thus, we have $5 x+3=-x$, so $x=-\frac{1}{2}$ is the solution.