Question

cho a,b,c là các số nguyên TM a+b=c^3+2024c

CM a^3+b^3+c^3⋮6

Answer 1

Ta có \(a+b=c^3+2024c\Leftrightarrow a+b-2024c=c^3\)

Thay vào, ta có \(a^3+b^3+c^3=a^3+b^3+2c^3-a-b-2024c=a^3-a+b^3-b+2c^3-2c-2022c=a\left(a-1\right)\left(a+1\right)+b\left(b-1\right)\left(b+1\right)+2c\left(c-1\right)\left(c+1\right)-2022c\)Lại có \(\left\{{}\begin{matrix}a\left(a-1\right)\left(a+1\right)⋮6\\b\left(b-1\right)\left(b+1\right)⋮6\\2c\left(c-1\right)\left(c+1\right)⋮6\\2022c⋮6\end{matrix}\right.\)

=> Ta có (đpcm)