Prove that the cube of any positive integer i | vedclass

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(A) Let $a$ be any positive integer. By Euclid's division lemma,we can express $a$ as $a = 3q + r$,where $r \in \{0, 1, 2\}$ and $q \ge 0$.Case $1$: I

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(A) Let $a$ be any positive integer. By Euclid's division lemma,we can express $a$ as $a = 3q + r$,where $r \in \{0, 1, 2\}$ and $q \ge 0$.Case $1$: If $r = 0$,then $a = 3q$. Cubing both sides,$a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m$,where $m = 3q^3$.Case $2$: If $r = 1$,then $a = 3q + 1$. Cubing both sides,$a^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 = 9m + 1$,where $m = 3q^3 + 3q^2 + q$.Case $3$: If $r = 2$,then $a = 3q + 2$. Cubing both sides,$a^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 = 9m + 8$,where $m = 3q^3 + 6q^2 + 4q$.Thus,the cube of any positive integer is of the form $9m$,$9m+1$,or $9m+8$.

Prove that the cube of any positive integer is of the form $9m$,$9m+1$,or $9m+8$,where $m$ is an integer.

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