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Find the cube root of each of the following cube numbers through estimation.
(i) \[19683\]
(ii) \[59319\]
(iii) \[85184\]
(iv) \[148877\]

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint: Here we have to determine the cube roots for all the given numbers. First we need to make the group of the three digits. After that for the left group we have to check where it lies. Then consider the number of the remaining three digits. Look at the third digit of the given number then you can say that the unit digit of its cube root must be in that third digit of the given number. So by using these conditions we can solve the given problem.

Complete step by step solution:
Let us look at the numbers one by one.
(i) \[19683\]: Making the groups of three-digits, it can be written as $(19)(683)$.
For the left-most group $(19)$, we observe that it lies between ${{2}^{3}}=8$ and ${{3}^{3}}=27$.
Therefore, considering the number of remaining groups of three-digits, the required cube root lies between \[20\] and \[30\].
Looking at the units digit \[3\] of the given number, we can say that the units digit of its cube root must be \[7\].
Since it is given that the number is definitely a cube, the cube root of \[19683\] is \[27\].
(ii) \[59319\]: Making the groups of three-digits, it can be written as $(59)(319)$.
For the left-most group $(59)$, we observe that it lies between ${{3}^{3}}=27$ and ${{4}^{3}}=64$.
Therefore, considering the number of remaining groups of three-digits, the required cube root lies between \[30\] and \[40\].
Looking at the units digit \[9\] of the given number, we can say that the units digit of its cube root must be \[9\].
Since it is given that the number is definitely a cube, the cube root of \[59319\] is \[39\].
(iii) \[85184\]: Making the groups of three-digits, it can be written as $(85)(184)$.
For the left-most group $(85)$, we observe that it lies between ${{4}^{3}}=64$ and ${{5}^{3}}=125$.
Therefore, considering the number of remaining groups of three-digits, the required cube root lies between \[40\] and \[50\].
Looking at the units digit \[4\] of the given number, we can say that the units digit of its cube root must be \[4\].
Since it is given that the number is definitely a cube, the cube root of \[85184\] is \[44\].
(iv) \[148877\]: Making the groups of three-digits, it can be written as $(148)(877)$.
For the left-most group $(148)$, we observe that it lies between ${{5}^{3}}=125$ and ${{6}^{3}}=216$.
Therefore, considering the number of remaining groups of three-digits, the required cube root lies between \[50\] and \[60\].
Looking at the units digit \[7\] of the given number, we can say that the units digit of its cube root must be \[3\].
Since it is given that the number is definitely a cube, the cube root of \[148877\] is \[57\].

Note: The above method works for perfect cubes only. e.g. \[19683\] and \[19783\] (etc.) both end in \[3\] but only \[19683\] is the perfect cube of \[27\].
Cubes of numbers from \[1\] to \[10\]:
${{1}^{3}}$\[001\]${{6}^{3}}$\[216\]
${{2}^{3}}$\[008\]${{7}^{3}}$\[343\]
${{3}^{3}}$\[027\]${{8}^{3}}$\[512\]
${{4}^{3}}$\[064\]${{9}^{3}}$\[729\]
${{5}^{3}}$\[125\]${{10}^{3}}$\[1000\]

Some other Properties of Cube Numbers:
If a number has \[n\] number of zeros at its end then it's cube will have \[3n\] number of zeros at its end.
The cube of an even number is always even and the cube of an odd number is always odd.
When a perfect cube is prime factorized, its factors can be grouped into triplets; groups of \[3\] identical primes.