pt $\Leftrightarrow (x+1)^3-8=3(\sqrt[3]{3x+5}-2)$$\Leftrightarrow(x-1)(x^2+4x+7)=3(\frac{3x+5-8}{\sqrt[3]{(3x+5)^2}+2\sqrt[3]{3x+5}+4})$
$\Leftrightarrow(x-1)(x^2+4x+7-\frac{9}{\sqrt[3]{(3x+5)^2}+2\sqrt[3]{3x+5}+4})=0$
$\color{red}{\Rightarrow x=1}$
$*$Ta có $-\frac{9}{\sqrt[3]{(3x+5)^2}+2\sqrt[3]{3x+5}+4}\geq -\frac{9}{3}=-3$
$\Rightarrow x^2+4x+7-\frac{9}{\sqrt[3]{(3x+5)^2}+2\sqrt[3]{3x+5}+4}\geq x^2+4x+4=(x+2)^2\geq 0$
Dấu $"="$ xảy ra khi $\begin{cases}x+2=0 \\ \sqrt[3]{3x+5}+1=0 \end{cases}\Rightarrow \left\{ \begin{array}{l} x=-2\\ x=-2 \end{array} \right.$
$\color{red}{\Rightarrow=-2} $