Sửa đề: \(a^3b-ab^3⋮6\)
\(a^3b-ab^3=ab\left(a^2-b^2\right)\)
\(=ab\left[\left(a^2-1\right)-\left(b^2-1\right)\right]\)
\(=ab\left(a^2-1\right)-ab\left(b^2-1\right)\)
\(=ab\left(a-1\right)\left(a+1\right)-ab\left(b-1\right)\left(a+1\right)\)
Vì \(a\left(a-1\right)\left(a+1\right);b\left(b-1\right)\left(b+1\right)\) là tích 3 số nguyên liên tiếp nên chúng chia hết cho cả 2 và 3
Mà \(\left(2;3\right)=1\)
\(\Rightarrow\left\{{}\begin{matrix}ab\left(a-1\right)\left(a+1\right)⋮6\\ab\left(b-1\right)\left(b+1\right)⋮6\end{matrix}\right.\)
\(\Rightarrow ab\left(a-1\right)\left(a+1\right)-ab\left(b-1\right)\left(b+1\right)⋮6\)
\(\Rightarrow a^3b-ab^3⋮6\)
