The original equation can be transformed into $(6 x+7)^{2}(6 x+8)(6 x+6)=72$.
Let $a=(6 x+7)^{2}, b=(6 x+8)(6 x+6)=(6 x+7)^{2}-1$.
Obviously, $a+(-b)=1, a \cdot(-b)=-72$.
Thus, we know that $a$ and $-b$ are the two distinct roots of the quadratic equation $y^{2}-y-72=0$, solving this quadratic equation yields $y=-8$ or $y=9$.
Therefore, $a=-8$ or 9, i.e., $(6 x+7)^{2}=-8$ (discard) or $(6 x+7)^{2}=9$.
From this, we obtain $x_{1}=-\frac{2}{3}$ or $x_{2}=-\frac{5}{3}$.
Example 11 Solve the equation $(6 x+7)^{2}(3 x+4)(x+1)=6$.

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