Question

Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
'The cube of a multiple of 7 is a multiple of 7³'.

Answer 1

Five multiples of 7 can be written by choosing different values of a natural number n in the expression 7n.

Let the five multiples be $7,14,21,28\text{and}35.$

The cubes of these numbers are:

$7^3=343,{14}^3=2744,{21}^3=9261,{28}^3=21952,\text{and}{35}^3=42875$

Now, write the above cubes as a multiple of 7³. Proceed as follows:

$343=7^3\times 1$
$2744={14}^3=14\times 14\times 14=\left(7\times 2\right)\times \left(7\times 2\right)\times \left(7\times 2\right)=\left(7\times 7\times 7\right)\times \left(2\times 2\times 2\right)=7^3\times 2^3$
$9261={21}^3=21\times 21\times 21=\left(7\times 3\right)\times \left(7\times 3\right)\times \left(7\times 3\right)=\left(7\times 7\times 7\right)\times \left(3\times 3\times 3\right)=7^3\times 3^3$
$21952={28}^3=28\times 28\times 28=\left(7\times 4\right)\times \left(7\times 4\right)\times \left(7\times 4\right)=\left(7\times 7\times 7\right)\times \left(4\times 4\times 4\right)=7^3\times 4^3$
$42875={35}^3=35\times 35\times 35=\left(7\times 5\right)\times \left(7\times 5\right)\times \left(7\times 5\right)=\left(7\times 7\times 7\right)\times \left(5\times 5\times 5\right)=7^3\times 5^3$

Hence, the cube of multiple of 7 is a multiple of 7³.