Đồ thị hàm số \(y = {x^3} - 3m{x^2} + 3mx + {m^2} - 2{m^3}\) tiếp xúc với trục hoành

\( \Leftrightarrow \) hệ phương trình \(\left\{ \begin{array}{l}
{x^3} - 3m{x^2} + 3mx + {m^2} - 2{m^3} = 0\\
3{x^2} - 6mx + 3m = 0
\end{array} \right.\) có nghiệm

\( \Leftrightarrow \left\{ \begin{array}{l}
{x^3} - 3m{x^2} + 3mx + {m^2} - 2{m^3} = 0\,\,\,\left( 1 \right)\\
{x^2} - 2mx + m = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\) 

(2) có nghiệm \( \Leftrightarrow \Delta ' = {m^2} - m \ge 0 \Leftrightarrow \left[ \begin{array}{l}
m \ge 1\\
m \le 0
\end{array} \right.\) 

(2) \( \Leftrightarrow {x^2} = m\left( {2x - 1} \right)\) 

TH1: \(x = \frac{1}{2} \Leftrightarrow \frac{1}{4} = 0\) (vô lí)

TH2: \(x \ne \frac{1}{2} \Rightarrow m = \frac{{{x^2}}}{{2x - 1}}\) 

Thay vào (1) ta có: \({x^3} - 3\frac{{{x^2}}}{{2x - 1}}{x^2} + 3\frac{{{x^2}}}{{2x - 1}}x + {\left( {\frac{{{x^2}}}{{2x - 1}}} \right)^2} - 2{\left( {\frac{{{x^2}}}{{2x - 1}}} \right)^3} = 0\) 

\(\begin{array}{l}
 \Leftrightarrow \frac{{{x^3}}}{{{{\left( {2x - 1} \right)}^3}}}\left[ {{{\left( {2x - 1} \right)}^3} - 3x{{\left( {2x - 1} \right)}^2} + 3{{\left( {2x - 1} \right)}^2} + x\left( {2x - 1} \right) - 2{x^3}} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
8{x^3} - 12{x^2} + 6x - 1 - 12{x^3} + 12{x^2} - 3x + 12{x^2} - 12x + 3 + 2{x^2} - x - 2{x^3} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
 - 6{x^3} + 14{x^2} - 10x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{1}{3} \Rightarrow S = 0 + \frac{1}{3} + 1 = \frac{4}{3}\\
x = 1
\end{array} \right.
\end{array}\)