pt ⇔(x+1)3−8=3(3√3x+5−2)⇔(x−1)(x2+4x+7)=3(3x+5−83√(3x+5)2+23√3x+5+4)
⇔(x−1)(x2+4x+7−93√(3x+5)2+23√3x+5+4)=0
⇒x=1
∗Ta có −93√(3x+5)2+23√3x+5+4≥−93=−3
⇒x2+4x+7−93√(3x+5)2+23√3x+5+4≥x2+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}