Question

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'

Answer 1

Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are:

$$\begin{gathered} 3^3=3\times 3\times 3=27 \\ {6}^3=6\times 6\times 6=216 \\ {9}^3=9\times 9\times 9=729 \\ {12}^3=12\times 12\times 12=1728 \\ {15}^3=15\times 15\times 15=3375 \end{gathered}$$

Now, let us write the cubes as a multiple of 27. We have:

$$\begin{gathered} 27=27\times 1 \\ 216=27\times 8 \\ 729=27\times 27 \\ 1728=27\times 64 \\ 3375=27\times 125 \end{gathered}$$

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.