Cho hàm số $f\left( x...

Ta có: $f'\left( x \right)=3{{x}^{2}}-4ax+{{a}^{2}}; f'\left( x \right)=0\Leftrightarrow 3{{x}^{2}}-4ax+{{a}^{2}}=0\left( * \right)$
Để hàm số có 2 điểm cực trị thì $\Delta '={{a}^{2}}>0\Leftrightarrow a\ne 0$
Khi đó (*) $\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{2a+a}{3}=a\Rightarrow y=b \\
& x=\dfrac{2a-a}{3}=\dfrac{a}{3}\Rightarrow y=\dfrac{4{{a}^{3}}}{27}+b \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& A\left( a;b \right) \\
& B\left( \dfrac{a}{3};\dfrac{4{{a}^{3}}}{27}+b \right) \\
\end{aligned} \right.$
Đặt $c=\dfrac{a}{3}\ne 0\Rightarrow \left\{ \begin{aligned}
& A\left( 3c;b \right) \\
& B\left( c;4{{c}^{3}}+b \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& A{{B}^{2}}=4{{c}^{2}}+16{{c}^{6}} \\
& O{{A}^{2}}=9{{c}^{2}}+{{b}^{2}} \\
& O{{B}^{2}}={{c}^{2}}+16{{c}^{6}}+8{{c}^{3}}b+{{b}^{2}} \\
\end{aligned} \right.$
Tam giác OAB vuông cân $\Leftrightarrow \left\{ \begin{aligned}
& O{{A}^{2}}=O{{B}^{2}} \\
& A{{B}^{2}}=2O{{A}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{c}^{2}}=2{{c}^{6}}+{{c}^{3}}b \\
& 8{{c}^{6}}=7{{c}^{2}}+{{b}^{2}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 2{{c}^{4}}+bc=1 \\
& 8{{c}^{6}}=7{{c}^{2}}+{{b}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& b=1;c=-1 \\
& b=-1;c=1 \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& b=1;a=-3 \\
& b=-1;a=3 \\
\end{aligned} \right.\Rightarrow T={{a}^{2}}+{{b}^{2}}=10$