Question

If $n$ be any positive integer, then by using Euclid's division algorithm, show that
(i) $n^3$ can be expressed in the form $9m,9m+1$ or $9m+8$.
(ii) $n^3+1$ can be expressed in the form $9m,9m+1$ or $9m+2$.
Here, $m$ is some integer.

Answer 1

Let $n$ be any positive integer and $b=3$

According to Euclid's division algorithm there exist integers $q$ and $r$ such that

$n=3\times q+r$ where $0\le r<3$

So value of $r$ can be $0,1$ and $2$

We can write

$n=3\times q$, $n=3\times q+1$ and $n=3\times q+2$

Now take the each case one by one

Case 1: $n=3\times q$

Take cube on both sides, we have

$n^3=27\times q^3$ = $9(3\times q^3)=9m$ where $m=3q^3$

$\therefore n^3+1=9m+1$

Similarly for 2nd case $n=3\times q+1$

Taking cube on both sides, we have

$n^3=27q^3+27q^2+9q+1$ or $n^3=9(3q^3+3q^2+q)+1$

Which is equal to $9m+1$ where $m=(3q^3+3q^2+q)$

Thus, $n^3+1=9m+2$

For final case $n=3\times q+2$

Taking cube on both sides, we have

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

i.e. $n^3=9(3q^3+6q^2+4q)+8$

which is equal to $9m+8$ where $m=(3q^3+6q^2+4q)$

Hence, $n^3+1=9m+8+1=9m+9=9(m+1)=9n$ where $n=m+1$ and $n$ is an integer