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# Write cubes of 5 natural numbers which are of form 3n+1 (e.g.4,7,10…) and verify the following;The cube of a natural number of the form 3n+1 is a natural number of form 3n+1.  Hint: We can find the five natural numbers of form 3n+1 by substituting 1,2,3,4,5 respectively in n and in this question 3n+1 implies that when we divide a number with 3 the remainder would be 1.

We have to write cubes of 5 natural numbers which are of form 3n+1 and we have to verify their cubes are natural numbers of form 3n+1.
The five natural numbers which are of form 3n+1 are 4,7,10,13 and 16 we obtain them by simply substituting 1,2,3,4,5 respectively in 3n+1.
The first thing we are going to do is writing cubes of 5 natural numbers (i.e. 4,7,10,13,17).
The cubes of 5 natural numbers (i.e. 4,7,10,13,16) are 64,343,1000,2197 and 4096 respectively.
The second thing we are going to do is expressing their cubes in the form 3n+1 if possible.
So, the cubes of natural numbers 4,7,10,13,17 can be expressed as:
$64=3\times 21+1$,64 is expressed as 3n+1 where n=21.
$343=3\times 114+1$, 343 is expressed as 3n+1 where n=114.
$1000=3\times 333+1$, 1000 is expressed as 3n+1 where n=333.
$2197=3\times 732+1$, 2197 is expressed as 3n+1 where n=732.
$4096=3\times 1365+1$, 4096 is expressed as 3n+1 where n=1365.
So, here we expressed cubes of numbers 4,7,10,13,16 of form 3n+1 in the form of 3n+1.
So, the statement “The cube of natural numbers of the form 3n+1 is a natural number of the form 3n+1.” is verified.

Note: Now we will verify the above question in an alternative way, we directly cube 3n+1. We use
${{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$ formula and proceed.
Now cubing 3n+1, ${{(3n+1)}^{3}}=27{{n}^{3}}+3{{n}^{2}}+3n+1$ , taking 3 common from the coefficients of ${{n}^{3}},{{n}^{2}}$ and $n$ .
Now it becomes,
$\Rightarrow 3(9{{n}^{3}}+{{n}^{2}}+n)+1\to (1)$
We assume $9{{n}^{3}}+{{n}^{2}}+n=k$
Equation $\text{(1)}$ becomes $3k+1$
Hence the statement “The cube of natural number of the form 3n+1 is a natural number of same
form.” is verified.
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