Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
'The cube of a natural number which is a multiple of 3 is a multiple of 27'
Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:
$3^{3}=3 \times 3 \times 3=27$
$6^{3}=6 \times 6 \times 6=216$
$9^{3}=9 \times 9 \times 9=729$
$12^{3}=12 \times 12 \times 12=1728$
$15^{3}=15 \times 15 \times 15=3375$
Now, let us write the cubes as a multiple of 27. We have:
$27=27 \times 1$
$216=27 \times 8$
$729=27 \times 27$
$1728=27 \times 64$
$3375=27 \times 125$
It is evident that the cubes of the above multiples of 3 could be written as multiples of 27. Thus, it is verified that the cube of a natural number, which is a multiple of 3, is a multiple of 27.
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