Question

Show that the cube of any positive integer is of the form $9m,9m+1$ or $9m+8$. Using Euclid's division lemma.

Answer 1

We have,

Let x be any positive integer.

Then,

It is of the form $3q,3q+1,3q+2$

Case 1:-

When $x=3q$

Then,

$x^3={\left(3q\right)}^3=27q^3=9\left(3q^3\right)=9mwhere,m=3q^3$

Case 2:-

When

$x=3q+1$

$Then,$

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

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

$=9q(3q^2+3q+1)+1$

$=9m+1,wherem=q(3q^2+3q+1)$

Case 3:-

When

$x=3q+2$

$Then,$

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

$=27q^3+54q^2+36q+8$

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

$x^3=9m+8$

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

Hence, this is the answer.