\(x_1^3-x_2^3=7\)
\(\Leftrightarrow\left(x_1-x_2\right)^3+3x_1x_2\left(x_1-x_2\right)=7\)
Do \(x_1-x_2=1\)
\(\Rightarrow1^3+3x_1x_2.1=7\)
\(\Rightarrow x_1x_2=2\)
\(\Rightarrow\left(x_1+x_2\right)^2=\left(x_1-x_2\right)^2+4x_1x_2=1^2+4.2=9\)
\(\Rightarrow\left[{}\begin{matrix}x_1+x_2=3\\x_1+x_2=-3\end{matrix}\right.\)
Theo hệ thức Viet:
TH1: \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-3\\b=2\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x_1+x_2=-3\\x_1x_2=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\)

