\(\Delta=m^2+4>0;\forall m\Rightarrow\) phương trình luôn có 2 nghiệm pb
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-1\end{matrix}\right.\)
\(x_1^3+x_2^3=-4\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=-4\)
\(\Leftrightarrow-m^3-3m=-4\)
\(\Leftrightarrow m^3+3m-4=0\)
\(\Leftrightarrow\left(m-1\right)\left(m^2+m+4\right)=0\)
\(\Leftrightarrow m=1\)
\(\Delta=m^2-4.1.\left(-1\right)=m^2+4>0\) suy ra pt luôn có 2 nghiệm phân biệt
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-1\end{matrix}\right.\)
\(x^3_1+x^3_2=-4\\ \Leftrightarrow\left(x_1+x_2\right)\left(x^2_1-x_1x_2+x_2^2\right)=-4\\ \Leftrightarrow-m\left[\left(x_1+x_2\right)^2-3x_1x_2\right]=-4\\ \Leftrightarrow m\left[\left(-m\right)^2-3.\left(-1\right)\right]=4\\ \Leftrightarrow m\left(m+3\right)-4=0\\ \Leftrightarrow m^2+3m-4=0\\ \Leftrightarrow m^2+4m-m-4=0\\ \Leftrightarrow m\left(m+4\right)-\left(m+4\right)=0\\ \Leftrightarrow\left(m+4\right)\left(m-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=-4\\m=1\end{matrix}\right.\)
