Áp dụng hằng đẳng thức :
\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\)
Ta viết lại phương trình thành :
\(\left(4x+3+5-7x+3x-8\right)\left[\left(4x+3\right)^2+\left(5-7x\right)^2+\left(3x-8\right)^2\right]+3\left(4x+3\right)\left(5-7x\right)\left(3x-8\right)=0\)
\(\Leftrightarrow3\left(4x+3\right)\left(5-7x\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\5-7x=0\\3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{5}{7}\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy : \(S=\left\{-\dfrac{3}{4};\dfrac{5}{7};\dfrac{8}{3}\right\}\).