Question

Show that the cube of a positive integer is of the form $9m$, $9m+1$, or $9m+8$.

Answer 1

Hint: We will rewrite the positive integer using Euclid's division lemma. The Euclid's division lemma states that given two positive integers $x$ and $q$, there exists whole numbers $p$ and $r$ such that $x=pq+r$ and $0\le r< p$. Then we will take the cube of the positive integer by assuming $p=3$. This will lead us to prove that the cube of a positive integer is of the form $9m$, $9m+1$ or $9m+8$.

Complete step-by-step solution
The Euclid's division lemma states that given two positive integers $x$ and $q$, there exists whole numbers $p$ and $r$ such that $x=pq+r$ and $0\le r< p$. So, let $x$ be a positive integer and let us assume that $p=3$. So, we have $x=3q+r$ and $0\le r< 3$. We have three possibilities for $r$ which are $r=0$, $r=1$ and $r=2$.
Let us look at the first case, which is $r=0$. So, the integer becomes $x=3q$. We will take the cube of this integer in the following manner,
$\begin{align}
  & {{x}^{3}}={{\left( 3q \right)}^{3}} \\
 & =27{{q}^{3}}
\end{align}$
This cube is of the form $9m$ where $m=3{{q}^{3}}$.
Now, we will look at the second case, that is $r=1$. So, the integer is $x=3q+1$. Taking the cube of this integer we get,
${{x}^{3}}={{\left( 3q+1 \right)}^{3}}$
Now, to expand the RHS of the above equation, we will use the following formula,
${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$
Using this formula we get,
${{x}^{3}}={{\left( 3q \right)}^{3}}+{{1}^{3}}+3\times 3q\times 1\times \left( 3q+1 \right)$
Simplifying the above expression, we have
$\begin{align}
  & {{x}^{3}}=27{{q}^{3}}+1+9q\left( 3q+1 \right) \\
 & =27{{q}^{3}}+1+27{{q}^{2}}+9q
\end{align}$
Rearranging the above equation and taking 9 as a common factor, we get
${{x}^{3}}=9\left( 3{{q}^{3}}+3{{q}^{2}}+q \right)+1$
We can see that the above expression is of the form $9m+1$ where $m=3{{q}^{3}}+3{{q}^{2}}+q$.
Next, we will look at the third case, which is $r=2$. SO, the integer will become $x=3q+2$. We will look at the cube of this integer as follows,
${{x}^{3}}={{\left( 3q+2 \right)}^{3}}$
We will again use the formula for the expansion of ${{\left( a+b \right)}^{3}}$ to expand the RHS of the above equation in the following manner,
${{x}^{3}}={{\left( 3q \right)}^{3}}+{{2}^{3}}+3\times 3q\times 2\times \left( 3q+2 \right)$
Simplifying the above equation we get,
$\begin{align}
  & {{x}^{3}}=27{{q}^{3}}+8+18q\left( 3q+2 \right) \\
 & =27{{q}^{3}}+54{{q}^{2}}+36q+8
\end{align}$
Taking 9 as a common factor, we get
${{x}^{3}}=9\left( 3{{q}^{3}}+6{{q}^{2}}+4q \right)+8$
We can see that the above expression is of the form $9m+8$ where $m=3{{q}^{3}}+6{{q}^{2}}+4q$.
Hence, we have proven that the cube of a positive integer is of the form $9m$, $9m+1$ or $9m+8$.

Note: It is essential that we are familiar with Euclid's division lemma. In the expression $x=pq+r$, $p$ is the divisor, $r$ is the remainder and $x$ is the dividend. In this type of question, it is useful to be familiar with algebraic identities involving cubes and squares of expressions. We should be careful while taking common factors from terms in an expression. It is always beneficial to write the calculations explicitly to avoid making minor mistakes.