`a^3+b^3=2(c^3-8d^3)`

`=>a^3+b^3+c^3+d^3= 2(c^3-8d^3)+c^3+d^3`

`=>a^3+b^3+c^3+d^3= 2c^3-16d^3+c^3+d^3`

`=>a^3+b^3+c^3+d^3=3c^3-15d^3`

`=>a^3+b^3+c^3+d^3=3(c^3-5d^3)`

`=>a^3+b^3+c^3+d^3` `vdots` `3`

Xét hiệu `:`

`a^3+b^3+c^3+d^3-(a+b+c+d)`

`=a^3-a+b^3-b+c^3-c+d^3-d`

`= a(a^2-1)+b(b^2-1)+c(c^2-1)+d(d^2-1)`

`= (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1)+(d-1)d(d+1)`

Ta luôn có:

`{((a-1)a(a+1) vdots 3),((b-1)b(b+1) vdots 3),((c-1)c(c+1) vdots 3):}`

`=> (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1)+(d-1)d(d+1)` `vdots` `3`

`=> a^3+b^3+c^3+d^3-(a+b+c+d)` `vdots` `3`

Mà `a^3+b^3+c^3+d^3` `vdots` `3(` cmt `)` nên `a+b+c+d` `vdots` `3(` dpcm `)`