Question

Show that every cube number is of the form $9n$ or $9n\pm 1$.

Answer 1

Hint: Generalize every cube number (as ${{a}^{3}}$), and use Euclid’s division lemma to obtain the proof. Expansion of algebraic formulas such as ${{(A+B)}^{3}}$is used.
${{(A+B)}^{3}}={{A}^{3}}+{{B}^{3}}+3{{A}^{2}}B+3A{{B}^{2}}$
Let $a$ be any positive integer. According to Euclid’s division lemma, $a=3q+r$, where $q\ge 0$(since we are considering positive integers ) and $3>r\ge 0$.
Every positive cube number can be represented using this form, that is, as ${{a}^{3}}$ or ${{(3q+r)}^{3}}$ . There exist three different cases, difference as in $r$ taking three different integral values between zero and three (zero inclusive).

Complete step by step answer:
Let us first consider case(1) at $r=0$.
At $r=0$,
$a=3q$
Taking cubes on both sides, we get
${{a}^{3}}=27{{q}^{3}}$
$=9(3{{q}^{3}})$(is of the form $9n$ where $3{{q}^{3}}=n$ )
$\therefore {{a}^{3}}=9n$
Now let us consider case(2) at $r=1$.
At $r=1$,
$a=3q+1$
Taking cubes on both sides, we have
${{a}^{3}}={{(3q+1)}^{3}}$
The RHS is in the form of the algebraic identity${{(A+B)}^{3}}$.
$\therefore {{a}^{3}}=27{{q}^{3}}+27{{q}^{2}}+9q+1$
$=9(3{{q}^{3}}+3{{q}^{2}}+q)+1$ (is of the form $9n+1$ where $3{{q}^{3}}+3{{q}^{2}}+q=n$ )
$\therefore {{a}^{3}}=9n+1$
Now let us consider case(3) at $r=2$.
At $r=2$,
$a=3q+2$
Taking cubes on both sides, we get
${{a}^{3}}={{(3q+2)}^{3}}$
The RHS is in the form of the algebraic identity${{(A+B)}^{3}}$.
${{a}^{3}}=27{{q}^{3}}+54{{q}^{2}}+36q+8$
$=27{{q}^{3}}+54{{q}^{2}}+36q+9-9+8$
$=9(3{{q}^{3}}+6{{q}^{2}}+4q+1)-1$ (is of the form $9n-1$ where $3{{q}^{3}}+6{{q}^{2}}+4q+1=n$ )
$\therefore {{a}^{3}}=9n-1$
Similarly proofs can be obtained for negative numbers as well.

Note:
 It is also possible to show that some cube numbers are of the form $9n+8$, in a similar way.
At $r=2$,
$a=3q+2$
Taking cubes on both sides,
${{a}^{3}}={{(3q+2)}^{3}}$
The RHS is in the form of the algebraic identity${{(A+B)}^{3}}$.
${{a}^{3}}=27{{q}^{3}}+54{{q}^{2}}+36q+8$
${{a}^{3}}=9(3{{q}^{3}}+6{{q}^{2}}+4q)+8$ (is of the form $9n+8$ where $3{{q}^{3}}+6{{q}^{2}}+4q=n$ )
$\therefore {{a}^{3}}=9n+8$

We use Euclid’s division lemma to obtain proof of this question. According to Euclid’s lemma, for two positive integers, $a$ and \[b\], there exists unique integers $q$ and $r$ which satisfies the condition $a=bq+r$, where $b>r\ge 0$.