I think it’s about the same level of difficulty as showing you need at least four, but yeah I agree that what is written is insufficient. One solution is that y

I think it’s about the same level of difficulty as showing you need at least four, but yeah I agree that what is written is insufficient. One solution is that you can write 2002 = 10^3 + 10^3 + 1^3 + 1^3 Then, multiply through by 2002^2001, which is itself a cube since 2001 is divisible by 3.