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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$$\Rightarrow x^2+4x+7-\frac{9}{\sqrt[3]{(3x+5)^2}+2\sqrt[3]{3x+5}+4}\geq x^2+4x+4=(x+2)^2=0$Dấu$"="$xảy ra khi$ $$\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} $

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} $