Question

An odd cube number will have a/an ____ cube root.
A. odd
B. even
C. fraction
D. None of these

Answer 1

Hint: Assume the cube root of an odd cube number is odd and check if the cube of an odd number is odd or not. If yes, the cube root of an odd number is odd else it is not i.e. the cube root of an odd number is an even. General form of an even number is $ 2n $ where n belongs to whole numbers and the general form of an odd number is $ 2n + 1 $ where n belongs to whole numbers.

Complete step-by-step answer:
An odd number can be written as $ 2n + 1 $
Assume the cube root of an odd cube number is an odd number.
 $ \sqrt[3]{{2m + 1}} = 2n + 1 $ Where m, n are whole numbers
We have to prove that $ {\left( {2n + 1} \right)^3} = 2m + 1 $
 $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2} $
 $
  {\left( {2n + 1} \right)^3} = {\left( {2n} \right)^3} + {1^3} + 3{\left( {2n} \right)^2}\left( 1 \right) + 3\left( {2n} \right){\left( 1 \right)^2} \\
   = 8{n^3} + 1 + 3 \times 4{n^2} + 6n \\
   = 8{n^3} + 1 + 12{n^2} + 6n \\
   = 8{n^3} + 12{n^2} + 6n + 1 \\
   = 2\left( {4{n^3} + 6{n^2} + 3n} \right) + 1 \\
   = 2m + 1 \\
  $
Where $ m = 4{n^3} + 6{n^2} + 3n $ can be any whole number.
Therefore, an odd cube number will have an odd cube root.
From among the options given in the question option A is correct, that is the cube root of an odd cube number is an odd number.
So, the correct answer is “Option A”.

Note: Another approach to solve the above problem
An even number can be written as $ 2n $
Assume the cube root of an odd cube number is an even number.
 $ \sqrt[3]{{2m + 1}} = 2n $ Where m, n are whole numbers
We have to prove that $ {\left( {2n} \right)^3} = 2m + 1 $
 $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2} $
 $
  {\left( {2n + 0} \right)^3} = {\left( {2n} \right)^3} + {0^3} + 3{\left( {2n} \right)^2}\left( 0 \right) + 3\left( {2n} \right){\left( 0 \right)^3} \\
   = 8{n^3} + 0 + 0 \\
   = 2\left( {4{n^3}} \right) \\
   = 2m \\
  $
Where $ m = 4{n^3} $ can be any whole number and $ {\left( {2n} \right)^3} \ne 2m + 1 $
We assumed that the cube root of an odd cube number is an even number which is wrong which means the cube root of an odd cube number is an odd number.