Đặt `k=x/2023=y/2024=z/2025`

`⇒x=2023k;y=2024k;z=2025k`

Ta có:

`(x-z)^3`

`=(2023k-2025k)^3`

`=[k(2023-2025)]^3`

`=(-2k)^3`

`=-8k^3`

Mặt khác:

`8(x-y)^2.(y-z)`

`=8.(2023k-2024k)^2.(2024k-2025k)`

`=8.[k(2023-2024)]^2.[k(2024-2025)]`

`=8.(-k)^2.(-k)`

`=8.(-k)^3`

`=-8k^3`

Vì `-8k^3=-8k^3`

Nên `(x-z)^3=8(x-y)^2.(y-z)`

Vậy `(x-z)^3=8(x-y)^2.(y-z)`          `text{(đpcm)}`