# Ex.1.1 Q5 Real Numbers Solution - NCERT Maths Class 10

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## Question

Use Euclid’s division lemma to show that the cube of any positive integer is of the form $$9\,\rm{m}, 9\,\rm{m} + 1$$ or $$9\,\rm{m} + 8.$$

Video Solution
Real Numbers
Ex 1.1 | Question 5

## Text Solution

To Prove:

The cube of any positive integer is of the form \begin{align} {9\,\rm{m}, 9\,\rm{m} + 1} \end{align} or \begin{align} {9\,\rm{m} + 8.} \end{align}

Reasoning:

Suppose that there is a positive integer ‘$$a$$’ . By Euclid’s lemma, we know that for positive integers $$a$$ and $$b,$$ there exist unique integers $$q$$ and $$r,$$ such that \begin{align} a=b q+r, 0 \leq r \lt b\end{align}

If we keep the value of $$b = 3$$, then $$0 ≤ r < 3$$  i.e. $$r = 0$$ or $$1$$ or $$2$$ but it can’t be $$3$$ because $$r$$ is smaller than $$3.$$ So, the possible values for $$a = 3q$$ or $$3q + 1$$ or $$3q + 2.$$ Now, find the cube of all the possible values of $$a.$$ If $$q$$ is any positive integer then its cube (let’s call it as “$$\rm{m}$$”) will also be a positive integer. Now, observe carefully that the cube of all the positive integers is either of the form $$9\,\rm{m}$$ or $$9\,\rm{m}+ 1$$ or $$9\,\rm{m} + 1$$ for some integer $$\rm{m}.$$

Steps:

Let \begin{align}{“a”} \end{align} be any positive integer and $$q=3.$$

Then, $$a=3q+r$$ for some integer $$q\ge 0$$ and $$0\le r<3$$

Therefore, $$a=3q$$ or $$3q+1$$ or $$3q+2$$

Case – I. When $$a=3q$$

${{\left( a \right)}^{3}}={{\left( 3q \right)}^{3~}}=27{{q}^{3}}=9{ }\left( 3{{q}^{3}} \right) =9\,\rm{m}$

Where $$\rm{m}$$ is an integer such that $$\text{m}=3{{q}^{3}}$$

Case – II. When $$a=3q+1$$

\begin{align}{{\left( a \right)}^{3}}&={ }{{\left( 3q{ }+{ }1 \right)}^{3}} \\{{\left( a \right)}^{3}}&={ }27{{q}^{3}}+{ }27{{q}^{2}}+9q{ }+1 \\ {{\left( a \right)}^{3}}&={ }9\left( 3{{q}^{3}}+{ }3{{q}^{2}}+q \right){ }+1 \\ {{\left( a \right)}^{3}}&={ }9\rm{m}+1 \\\end{align}

Where $$\rm{m}$$ is an integer such that $$\text{m}=3{{q}^{3}}+{ }3{{q}^{2}}+{ }q$$

Case – III. When $$a=3q+2$$

\begin{align}\,{{\left( a \right)}^{3}}&={ }{{\left( 3q+2 \right)}^{3}} \\ {{\left( a \right)}^{3}}&={ }27{{q}^{3}}+{ }54{{q}^{2}}+36q{ }+{ }8 \\ {{\left( a \right)}^{3}}&={ }9\left( 3{{q}^{3}}+{ }6{{q}^{2}}+{ }4q \right){ }+{ }8 \\ {{\left( a \right)}^{3}}&={ }9\rm{m}+8 \\\end{align}

Where $$\rm{m}$$ is an integer such that $$\text{m}=3{{q}^{3}}+{ }6{{q}^{2}}+{ }4q$$

Thus, we can see that the cube of any positive integer is of the form $$9\rm{m},\; 9{m}+ 1$$ or $$9\rm{m} +8.$$

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