tiếp cho bài pt nãy à :v
Từ \(a+b=4\Rightarrow a=4-b\)
\(\Rightarrow\left(4-b\right)^3+b^3=28\)
\(\Rightarrow-b^3+12b^2-48b+64+b^3-28=0\)
\(\Rightarrow12b^2-48b+36=0\)
\(\Rightarrow12\left(b-1\right)\left(b-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=1\\b=3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=4-b=4-1=3\\a=4-b=4-3=1\end{matrix}\right.\)