\(( 5 x + 1 ) ^3 − ( 3 x + 2 )^3 − ( 2 x − 1 ) ^3 = 0\)
`=>3x+2=a;2x-1=b`
\(⇒ a + b = 3 x + 2 + 2 x − 1 = 5 x + 1\)
Thay vào `(1)` :
\(( a + b ) ^ 3-a^3-b^3=0\)
`=>a^3 + 3a^2b + 3ab^2 + b^3 - a^3 - b^3 = 0`
\(⇒ 3 a b ( a + b ) = 0\)
\(\Rightarrow\left[{}\begin{matrix}3ab=0\\a+b=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}ab=0\\a+b=0\end{matrix}\right.\)
Với `ab=0`
\(⇒ ( 3 x + 2 ) ( 2 x − 1 ) = 0\)
\(\Rightarrow\left[{}\begin{matrix}3x+2=0\\2x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=-2\\2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Với `a+b=0`
\(⇒ 3 x + 2 + 2 x − 1 = 0\)
\(⇒ 5 x = − 1\)
\(\Rightarrow x=-\dfrac{1}{5}\)
\(x\in\left\{-\dfrac{2}{3};\dfrac{1}{2};-\dfrac{1}{5}\right\}\)
