Ta có \(A=a^5b-ab^5=a^5b-ab-ab^5+ab\)
\(A=\left(a^5b-ab\right)-\left(ab^5-ab\right)\)
\(A=b\left(a^5-a\right)-a\left(b^5-b\right)\)
Ta có \(m^5-m=m\left(m^4-1\right)=m\left(m^2-1\right)\left(m^2+1\right)\)
\(=m\left(m+1\right)\left(m-1\right)\left(m^2-4+5\right)\)
\(=m\left(m-1\right)\left(m+1\right)\left(m^2-4\right)-5m\left(m-1\right)\left(m+1\right)\)
\(=m\left(m-1\right)\left(m+1\right)\left(m-2\right)\left(m+2\right)-5m\left(m-1\right)\left(m+1\right)\)
\(=\left(m-2\right)\left(m-1\right)m\left(m+1\right)\left(m+2\right)-5\left(m-1\right)m\left(m+1\right)\)
Vì \(m-2;m-1;m;m+1;m+2\) là 5 số nguyên liên tiếp nên chia hết cho 2 ; 3 ; 5
Mà \(\left(2;3;5\right)=1\)
\(\Rightarrow\left(m-2\right)\left(m-1\right)m\left(m+1\right)\left(m+2\right)\) chia hết cho \(2\times3\times5=30\)
\(\Rightarrow m^5-m\) chia hết cho 30
\(\Rightarrow a^5-a\) và \(b^5-b\) Chia hết cho 30
\(\Rightarrow b\left(a^5-a\right)-a\left(b^5-b\right)\) chia hết cho 30
\(\Rightarrow A=a^5b-ab^5\) chia hết cho 30
Vậy A chia hết cho 30
