Prove by contradiction that if $n^3$ is a multiple of $3$, then $n$ is a multiple of $3$ Problem statement:
Using proof by contradiction, prove that if $n^3$ is a multiple of $3$ , then $n$ is a multiple of $3.$
Attempt 1:
Assume that there is exist $n$ which  is a multiple of $3$ such that $n^3$ is not a multiple of $3.$
Then $n = 3k $ , $n^3 = 27 k^3 $ which a multiple of $3$, which contradicts the assumption that $n^3$ is a multiple of $3.$
Attempt 2:
Assume that there exist $n^3$ which is a multiple of $3$ such that $n$ is not a multiple of $3.$
Then $n = k+ 1 $ or $n= k+2.$
Then $n^3 = k^3 + 3k^2 + 3k + 1$
or $n^3 = k^3 + 6 k^2 + 12 k + 8.$
Both cases contradict the assumption that $n^3 $ is a multiple of $3.$
Which solution is correct ? Or do both work ?

given statement: “If $n^3$ is a multiple of $3,$ then $n$ is a multiple of $3.$”

*

*Your Attempt 1 is wrong, because it's attempting to prove the
converse of the given statement.

*Attempt 2 (“Assume (to the contrary) that there exists $n^3$ which
is a multiple of $3$ such that $n$ is not a multiple of $3$”) does start correctly.  However, the correct translation of <$n^3$ is a multiple of $3$> ought to be <$n^3=3k+1\,$ or $\,3k+2$> instead.

*Alternatively and equivalently, you could also start with “Suppose
that $n^3$ is a multiple of $3,$ and assume (to the contrary) that
that $n$ is not a multiple of $3.$”

