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

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Hint: Any number can be written in the form of $3q$or of $3q + 1$or $3q + 2$. Find the cube of all of them making different cases for each.

Let $x$be any positive integer. Then $x$will be either of the form of $3q$or of $3q + 1$or $3q + 2$. So, we have the following cases:

Case 1: When $x = 3q$

In this case, we know:

$

\Rightarrow {x^3} = {\left( {3q} \right)^3} = 27{q^3}, \\

\Rightarrow {x^3} = 9\left( {3{q^3}} \right) = 9m,{\text{ where }}m = 3{q^3}{\text{ }} \\

$

Case 2: When $x = 3q + 1$

In this case we have:

\[

\Rightarrow {x^3} = {\left( {3q + 1} \right)^3}, \\

\Rightarrow {x^3} = 27{q^3} + 27{q^2} + 9q + 1, \\

\Rightarrow {x^3} = 9q\left( {3{q^2} + 3q + 1} \right) + 1, \\

\Rightarrow {x^3} = 9m + 1,{\text{ where }}m = q\left( {3{q^2} + 3q + 1} \right) \\

\]

Case 3: When $x = 3q + 2$

In this case we have:

\[

\Rightarrow {x^3} = {\left( {3q + 2} \right)^3}, \\

\Rightarrow {x^3} = 27{q^3} + 54{q^2} + 36q + 8, \\

\Rightarrow {x^3} = 9q\left( {3{q^2} + 6q + 4} \right) + 8, \\

\Rightarrow {x^3} = 9m + 8,{\text{ where }}m = q\left( {3{q^2} + 6q + 4} \right) \\

\]

Thus, ${x^3}$can be either of the form of $9m,9m + 1$ or $9m + 8$.

Note: From the above solution, we can say that the cube of any natural number can be written in the form of either $9m,9m + 1$ or $9m + 8$. From this we can conclude that when a cube of any natural number is divided by 9, it gives remainder 0, 1 or 8.

Let $x$be any positive integer. Then $x$will be either of the form of $3q$or of $3q + 1$or $3q + 2$. So, we have the following cases:

Case 1: When $x = 3q$

In this case, we know:

$

\Rightarrow {x^3} = {\left( {3q} \right)^3} = 27{q^3}, \\

\Rightarrow {x^3} = 9\left( {3{q^3}} \right) = 9m,{\text{ where }}m = 3{q^3}{\text{ }} \\

$

Case 2: When $x = 3q + 1$

In this case we have:

\[

\Rightarrow {x^3} = {\left( {3q + 1} \right)^3}, \\

\Rightarrow {x^3} = 27{q^3} + 27{q^2} + 9q + 1, \\

\Rightarrow {x^3} = 9q\left( {3{q^2} + 3q + 1} \right) + 1, \\

\Rightarrow {x^3} = 9m + 1,{\text{ where }}m = q\left( {3{q^2} + 3q + 1} \right) \\

\]

Case 3: When $x = 3q + 2$

In this case we have:

\[

\Rightarrow {x^3} = {\left( {3q + 2} \right)^3}, \\

\Rightarrow {x^3} = 27{q^3} + 54{q^2} + 36q + 8, \\

\Rightarrow {x^3} = 9q\left( {3{q^2} + 6q + 4} \right) + 8, \\

\Rightarrow {x^3} = 9m + 8,{\text{ where }}m = q\left( {3{q^2} + 6q + 4} \right) \\

\]

Thus, ${x^3}$can be either of the form of $9m,9m + 1$ or $9m + 8$.

Note: From the above solution, we can say that the cube of any natural number can be written in the form of either $9m,9m + 1$ or $9m + 8$. From this we can conclude that when a cube of any natural number is divided by 9, it gives remainder 0, 1 or 8.

Last updated date: 01st Jun 2023

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