Let $X =\{n^3 + 3n^2 +3n | n>0\}$ and $Y = \{ n^3 -1 | n>0\}.$ Prove that $X =Y$.

Question

Let $X=\{n^3+3n^2+3n\mid n>0\},$ and $Y=\{n^3-1\mid n>0 \}$. Prove that $X=Y$.

I tried proving it by adding some values to $X$ and $Y$, but I could not make it equal.

Answer 1

It is not true. Presumably $n$ ranges over the natural numbers. $Y$ includes $0$ from $n=1$ while $X$ does not. Otherwise the sets are equal because the expression for $X$ is $(n+1)^3-1$, so every element that is in $X$ will be in $Y$. It is made harder to see by the fact that $n$ is used to define both $X$ and $Y$, but it is only a dummy variable. Make the definition $Y=\{m^3-1|m\gt 0\}$ and you can say the element of $X$ that is caused by $n$ is also an element of $Y$ that is caused by $m=n+1$.

The statement would be true if the definition of $X$ were changed to include $n \ge 0$