Ta có: \(8x^3+2x=\sqrt[3]{x+7}+x+7\)
Đặt \(\sqrt[3]{x+7}=t\)
\(\Rightarrow8x^3+2x=t+t^3\)
\(\Leftrightarrow\left(2x-t\right)\left(4x^2+2xt+t^2\right)+\left(2x-t\right)=0\)
\(\Leftrightarrow\left(2x-t\right)\left(4x^2+2xt+t^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=t\\4x^2+2xt+t^2+1=0\end{matrix}\right.\)
Với 2x=t \(\Leftrightarrow2x=\sqrt[3]{x+7}\Leftrightarrow8x^3-x-7=0\)
\(\Leftrightarrow\left(x-1\right)\left(8x^2+8x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\8x^2+8x+7=0\left(loại\right)\end{matrix}\right.\)
Với \(4x^2+2xt+t^2+1=0\)
Do \(4x^2+2xt+t^2+1=\left(x+t\right)^2+3x^2+1\ge1>0\)
⇒ ptvn
