Question

A natural number is said to be a perfect cube if it is the cube of a:
A.Real number
B.Integer
C.Natural number
D.None of these

Answer 1

Hint: A natural number (n) is said to be a perfect cube if (n = m³) i.e. it is the cube of some natural number or it means its cube root is a natural number $ \left( {\sqrt[3]{n} = m} \right) $ .

Complete step-by-step answer:
What is a cube of a number?
When a number is multiplied three times by itself, the product obtained is called the cube of a number.
For a given number ‘m’ , we define, cube of m = m × m × m, denoted by m³.
For example:
Thus, the cube of 2 is 8.
Thus, the cube of 3 is 27.
\[\left( {iii} \right){\text{ }}4{\text{ }} \times {\text{ }}4{\text{ }} \times {\text{ }}4{\text{ }} = {\text{ }}64\], here 64 is the cube of 4.
Perfect cube:
A natural number (n) is said to be a perfect cube if (n = m³) i.e. it is the cube of some natural number or this means its cube root is a natural number $ \left( {\sqrt[3]{n} = m} \right) $ .
For example:
Thus 1, 8, 27, 64, 125, etc. are perfect cubes.

Note: A given natural number is a perfect cube if it can be expressed as the product of triplets of equal factors.
How to find a number is perfect cube or not:
Every number can be expressed as the product of the power of its Prime factors. If the power of all the Prime factors is in the multiple of 3, then the number is said to be a perfect cube.
For example:
 $ \begin{gathered}
  1)15 = 3 \times 5 \\
   \Rightarrow {15^3} = 15 \times 15 \times 15 = \left( {3 \times 5} \right) \times \left( {3 \times 5} \right) \times \left( {3 \times 5} \right) \\
   \Rightarrow {15^3} = \left( {3 \times 3 \times 3} \right) \times \left( {5 \times 5 \times 5} \right) = {3^3} \times {5^3} \\
\end{gathered} $
From the above example, we can observe that each prime factor of a number appears three times in the prime factorization of its cube.
Hence, 15 is a perfect cube.
Let us write down, stating the prime factors of the number 600.
$\begin{gathered}
  600 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 \\
   \Rightarrow 600 = {2^3} \times 3 \times {5^2} \\
\end{gathered} $
The number 600 is not a perfect cube as all the prime factors is not a multiple of three.