a) \(A=a^3b-ab^3=\left(a^3b-ab\right)-\left(ab^3-ab\right)\)
\(=b.a\left(a^2-1\right)-a\left(b^3-b\right)\)
\(=a\left(a-1\right)\left(a+1\right)b-a\left(b-1\right)b\left(b+1\right)\)
\(Do:\)\(a-1\) \(;\)\(a\) \(;\) \(a+1\) là 3 số liên tiếp nên :
\(\left(a-1\right)a\left(a+1\right)\) \(⋮6\)
Tương tự : \(\left(b-1\right)b\left(b+1\right)\) \(⋮6\)
\(\Rightarrow\) \(A\) \(⋮\)\(6\)
\(a^5b-ab^5=a^5b-ab\left(ab^5-ab\right)\)
\(=b\left(a^5-a\right)-a\left(b^5-b\right)\)
Xét : \(m^5-m=m\left(m^4-1\right)=m\left(m^2-1\right)\left(m^2+1\right)\)
\(=m\left(m^2-1\right)\left(m+1\right)\left(m^2+4-5\right)\)
\(=m\left(m-1\right)\left(m+1\right)\left(m^2-4\right)+5\left(m-1\right)\left(m+1\right)\)
\(=\left(m-2\right)\left(m-1\right)m\left(m+1\right)\left(m+2\right)+5\left(m-1\right)m\left(m+1\right)\)
Ta thấy : \(\left(m-2\right);\left(m-1\right);m;\left(m+1\right);\left(m+2\right)\) là 5 số tự nhiên liên tiếp nên :
\(\left(m-2\right)\left(m-1\right)m\left(m+1\right)\left(m+2\right)\) \(⋮\) \(6\) \(\left(1\right)\)
\(m\left(m-1\right)\left(m+1\right)\) \(⋮\)\(6\)
\(5⋮5\)
\(\Rightarrow\) \(5m\left(m-1\right)\left(m+1\right)\) \(⋮\)\(30\) \(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra \(\left(m^5-m\right)\) \(⋮\)\(30\)
\(\Rightarrow\)\(\hept{\begin{cases}b\left(a^5-a\right)⋮30\Rightarrow\left(a^5b-ab^5\right)⋮30\\a\left(b^5-b\right)⋮30\end{cases}}\)
