Question

Which of the following numbers are not perfect cubes?
A) 64
B) 729
C) 243
D) 81

Answer 1

Hint: Here we may check each of the numbers provided in the options whether they are perfect cubes or not. To check we may use here the prime factorization method. So, we will break the given numbers into products of prime factors.

Complete step-by-step answer:
Using prime factorization let us break all the given numbers into their respective prime factors and check whether they are perfect cubes or not.
First of all let us check the number 64 that is the option (a):
So, we can write 64 in terms of its prime factors as:
$64=2\times 2\times 2\times 2\times 2\times 2$
We can also write it as:
$\begin{align}
  & 64=4\times 4\times 4 \\
 & 64={{\left( 4 \right)}^{3}} \\
\end{align}$
So, we can see that 64 is the cube of 4.
Now, let us check the option (b) :
We can write the number 729 in terms of its prime factors as:
$729=3\times 3\times 3\times 3\times 3\times 3$
We can also write it as:
$\begin{align}
  & 729=9\times 9\times 9 \\
 & 729={{\left( 9 \right)}^{3}} \\
\end{align}$
So, we can see that 729 Is the cube of 9.
Now, let us check the option (c):
We can write the number 243 in terms of its prime factors as:
$243=3\times 3\times 3\ \times 3\times 3$
We can also write it as:
$243={{\left( 3 \right)}^{3}}\times 3\times 3$
So, we can see that 243 is not a perfect cube.
Now, let us check the option (d):
We can write the number 81 in terms of its prime factors as:
$81=3\times 3\times 3\times 3$
We can also write it as:
$81={{\left( 3 \right)}^{3}}\times 3$
So, we can see that 81 is also not a perfect cube:
Hence, option (c) 243 and option (d) 81 are correct options.

Note: Students should note here that while finding the cube roots using the prime factorization method we should always try to make a group of three same prime factors.