\(\Delta'=m^2-m^2+2=2>0\)
Phương trình luôn có 2 nghiệm phân biệt
\(\left|x_1-x_2\right|=\left|\frac{2\sqrt{\Delta'}}{a}\right|=2\sqrt{2}\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2-2\end{matrix}\right.\)
\(\left|x_1^3-x_2^3\right|=10\sqrt{2}\)
\(\Leftrightarrow\left|\left(x_1-x_2\right)\left(x_1^2+x_2^2+x_1x_2\right)\right|=10\sqrt{2}\)
\(\Leftrightarrow\left|x_1-x_2\right|\left(x_1^2+x_1x_2+x_2^2\right)=10\sqrt{2}\)
\(\Leftrightarrow x_1^2+x_1x_2+x_2^2=5\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-x_1x_2=5\)
\(\Leftrightarrow4m^2-\left(m^2-2\right)=5\)
\(\Leftrightarrow m^2=1\Rightarrow m=\pm1\)