\(\Leftrightarrow x^3-6x^2+11x-m=0\) (1) có 3 nghiệm pb \(x=\left\{a;b;c\right\}\)
Theo định lý Viet:
\(\left\{{}\begin{matrix}a+b+c=6\\ab+bc+ca=11\\abc=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2b+b=6\\b\left(a+c\right)+ac=11\\abc=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=2\\2b^2+ac=11\\m=abc\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=2\\ac=11-2b^2=3\\m=b.ac=2.3=6\end{matrix}\right.\)
Vậy \(m=6\)