Đặt \(\sqrt[3]{x}=a;\sqrt[3]{2x-3}=b\)
=> \(4a^3+4b^3=4x+4\left(2x-3\right)=12x-12=12\left(x-1\right)=\left(a+b\right)^3\)
<=> \(4\left(a^3+b^3\right)=\left(a+b\right)^3\Leftrightarrow4\left[\left(a+b\right)^3-3ab\left(a+b\right)\right]-\left(a+b\right)^3=0\)
<=> \(4\left(a+b\right)^3-12ab\left(a+b\right)-\left(a+b\right)^3=0\)
<=> \(3\left(a+b\right)^3-12ab\left(a+b\right)=0\Leftrightarrow3\left(a+b\right)\left[\left(a+b\right)^2-4ab\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=0\)
<=> a = -b hoặc a = b