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 $7^3$′.

Given:

The cube of a multiple of 7 is a multiple of $7^3$

To do:

We have to write the cubes of 5 natural numbers of which are multiples of 7 and verify the given statement.

Solution: â€Š

First 5 natural numbers which are multiple of 7 are 7, 14, 21, 28, 35.

Therefore,

$(7)^3=7\times7\times7=343$

$(14)^3=(2\times7)^3=2^3\times7^3$

$(21)^3=(3\times7)^3=3^3\times7^3$

$(28)^3=(4\times7)^3=4^3\times7^3$

$(35)^3=(5\times7)^3=5^3\times7^3$

Any number multiplied by $7^3$ is a multiple of it.

Therefore,

$7^3, 14^3, 21^3, 28^3, 35^3$ are all of multiples of $7^3$

Hence, the given statement is true.