Question

Cho 3 số tự nhiên a,b,c. C/m: 

Nếu a+b+c chia hết cho 3 thì a³+b³+c³+3a²+3b²+3c² chia hết cho 6

Answer 1

Ta có: \(S=a^3+b^3+c^3+3a^2+3b^2+3c^2\)

\(=\left(a^3-a\right)+\left(b^3-b\right)+\left(c^3-c\right)+\left(3a^2-3a\right)+\left(3b^2-3b\right)+\left(3c^2-3c\right)+4\left(a+b+c\right)\)

\(=a\left(a+1\right)\left(a-1\right)+b\left(b-1\right)\left(b+1\right)+c\left(c-1\right)\left(c+1\right)+3a\left(a-1\right)+3b\left(b-1\right)+3c\left(c-1\right)+4\left(a+b+c\right)\)

Ta thấy: \(\hept{\begin{cases}a\left(a-1\right)\left(a+1\right)⋮6\\b\left(b-1\right)\left(b+1\right)⋮6\\c\left(c-1\right)\left(c+1\right)⋮6\end{cases}}\)(1)

\(\hept{\begin{cases}3a\left(a-1\right)⋮6\\3b\left(b-1\right)⋮6\\3c\left(c-1\right)⋮6\end{cases}}\)(2)

\(4\left(a+b+c\right)⋮6\)(3)

Từ (1),(2),(3) ta suy ra \(S⋮6\)