Which of the following are the perfect cubes?
A. $729$
B. $10000$
C. $1729$
D. none of these

Hint: To obtain which of the given numbers is a perfect cube we will use prime factorization method. Firstly we will find the prime factor of the number and then we will check whether we are getting a product of all the numbers multiplied by itself three times. If yes it is a perfect cube otherwise not.

Complete step-by-step answer:
We will check for each option given as follows:
Option (A)
 We have to whether the below number is a perfect cube:
$729$
So we will find the prime factor of the number as follows:
$\begin{align}
  & 3\left| \!{\underline {\,
  729 \,}} \right. \\
 & 3\left| \!{\underline {\,
  243 \,}} \right. \\
 & 3\left| \!{\underline {\,
  81 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1 \\
\end{align}$
So we get the prime factors as:
$729=3\times 3\times 3\times 3\times 3\times 3$
We can rewrite it as:
$729={{3}^{3}}\times {{3}^{3}}$
We are getting two terms which are multiplied by it-self three times so:
$729={{9}^{3}}$
So we get that 729 is a perfect cube.
Option (B)
We have to whether the below number is a perfect cube:
$10000$
So we will find the prime factor of the number as follows:
$\begin{align}
  & 2\left| \!{\underline {\,
  10000 \,}} \right. \\
 & 2\left| \!{\underline {\,
  5000 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2500 \,}} \right. \\
 & 2\left| \!{\underline {\,
  1250 \,}} \right. \\
 & 5\left| \!{\underline {\,
  625 \,}} \right. \\
 & 5\left| \!{\underline {\,
  125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & 1 \\
\end{align}$
So we get the prime factors as:
$10000=2\times 2\times 2\times 2\times 5\times 5\times 5\times 5$
We can rewrite it as:
$10000={{2}^{4}}\times {{5}^{4}}$
We are getting two terms multiplied four times by it-self.
So we get that 10000 is not a perfect cube.
Option (C)
 We have to whether the below number is a perfect cube:
$1729$
So we will find the prime factor of the number as follows:
$\begin{align}
  & 7\left| \!{\underline {\,
  1729 \,}} \right. \\
 & 13\left| \!{\underline {\,
  247 \,}} \right. \\
 & 19\left| \!{\underline {\,
  19 \,}} \right. \\
 & 1 \\
\end{align}$
So we get the prime factors as:
$1729=7\times 13\times 19$
We are not getting cube of any term
So we get that 1729 is not a perfect cube.

So, the correct answer is “Option A”.

Note: Factors of any number is a number which divide it completely. When we are talking about prime factorization we involve only prime numbers as a factor. This method is widely used to find the square root or cube root of any number. When we multiply a number by it-self three times the number obtained is known as cube of that number.