Question

Use Euclid's Division Lemma to show that the cube of any positive integer is of the form $9m,9m+1$ or $9m+8$, for some integer $m$.

Answer 1

Let $x$ be any positive integer. Then, it is of the form $3q$ or,$3q+1$ or, $3q+2$.

Case-I When $x=3q$

$\Rightarrow x^3=(3q{)}^3=27q^3=9(3q{3}^{)}=9m,$ where $m=3q^3$

Case-II when $x=3q+1$

$\Rightarrow x^3=(3q+1{)}^3$

$\Rightarrow x^3=27q^3+27q^2+9q+1$

$\Rightarrow x^3=9q(3q^2+3q+1)+1$

$\Rightarrow x^3=9m+1$, where $m=q(3q^2+3q+1)$

Case-III when $x=3q+2$

$\Rightarrow x^3=(3q+2{)}^3$

$\Rightarrow x^3=27q^3+54q^2+36q+8$

$\Rightarrow x^3=9q(3q^2+6q+4)+8$

$\Rightarrow x^3=9m+8$,where $m=q(3q^2+6q+4)$

Hence, $x^3$ is either of form $9m$ or $9m+1$ or $9m+8$.