Điều kiện: \(108x^3+12x\ge0\)
\(\Leftrightarrow x\ge0\)
Đặt \(3x=a\ge0\) thì ta có:
\(a^4+5=3\sqrt[3]{4a^3+4a}\)
\(\Leftrightarrow a^4-1=3\left(\sqrt[3]{4a^3+4a}-2\right)\)
\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\left(a^2+1\right)=\dfrac{12\left(a^3+a-2\right)}{\sqrt[3]{\left(4a^2+4a\right)^2}+2\sqrt[3]{\left(4a^2+4a\right)}+4}\)
\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\left(a^2+1\right)-\dfrac{12\left(a-1\right)\left(a^2+a+2\right)}{\sqrt[3]{\left(4a^2+4a\right)^2}+2\sqrt[3]{\left(4a^2+4a\right)}+4}=0\)
\(\Leftrightarrow\left(a-1\right)\left(\left(a+1\right)\left(a^2+1\right)-\dfrac{12\left(a^2+a+2\right)}{\sqrt[3]{\left(4a^2+4a\right)^2}+2\sqrt[3]{\left(4a^2+4a\right)}+4}\right)=0\)
\(\Leftrightarrow a=1\)
\(\Rightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
