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  \[7, 14, 21, 28 \text{ and }  35 .\]

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.