Question

You are told that 1331 is a perfect square. Can you guess without factorisation, what is its cube root? Similarly, guess the cube root of 4913, 12167, 32768.

Answer 1

Hint: Here we will use the concept of dividing into groups of three from the rightmost and check for the unit's place of required cube root from the group ,Similarly check for tenth’s place of required cube root from the group then we will get the required cube root of the number.

Complete step-by-step answer:
The given number is 1331. Let’s form the groups of three (because of cube root) digits starting from the rightmost digit. Therefore, the two groups are 1 and 331.
Consider the group 331: We take the unit place of required cube root as 1. Since, the unit place of the 331 is 1 and ${1^3} = 1$.
Consider the group 1: The unit place of 1 is 1 itself. So, we take 1 as ten’s place of cube root of 1331 as 1.
Therefore, $\sqrt[3]{{1331}} = 11$ .

Similarly, for 4913, the groups will be 4 and 913.
Consider the group 913: Here the unit place of the 913 is 3, so the unit place of the cube root will also be 3. The unit place of any cube root will be 3 if it ends with $7({7^3} = 343,{27^3} = 19683)$. Hence the cube root of 4913 will have unit place as 7.
Cube root of 4913 is $\_7$.
Now consider the group 4: we know that ${1^3} = 1$ and ${2^3} = 8$.
So,
 $
  1 < 4 < 8 \\
  {1^3} < 4 < {2^3} \\
$
We have to take the smallest number for the ten’s digit of the cube root i.e.., 1
Therefore, the cube root of 4913 is 17 i.e. $\sqrt[3]{{4913}} = 17$.

Similarly, for 12167, the groups will be 12 and 167.
Consider the group 167: Here the unit place of the 167 is 7, so the unit place of the cube root will also be 7. The unit place of any cube root will be 7 if it ends with $3({3^3} = 27,{13^3} = 2797)$. Hence the cube root of 12167 will have unit place as 3.
Hence, the cube root of 12167 is _3.
Now consider the group 12: we know that ${2^3} = 8$ and ${3^3} = 27$.
So,
 $
  8 < 12 < 27 \\
  {2^3} < 12 < {3^3} \\
 $
We have to take the smallest number for the ten’s digit of the cube root i.e.., 2
Therefore, the cube root of 12167 is 23 i.e. $\sqrt[3]{{12167}} = 23$.

Similarly, for 32768, the groups will be 32 and 768.
Consider the group 768: Here the unit place of the 768 is 8, so the unit place of the cube root will also be 8. The unit place of any cube root will be 8 if it ends with $2({2^3} = 8,{12^3} = 1728)$. Hence the cube root of 32768 will have unit place as 2.
Hence, Cube root of 32768 is _2.
Now consider the group 32: we know that ${3^3} = 27$and${4^3} = 64$.
So,
 $
  27 < 32 < 64 \\
  {3^3} < 32 < {4^3} \\
 $
We have to take the smallest number for the ten’s digit of the cube root i.e.., 3
Therefore, the cube root of 32768 is 32 i.e. $\sqrt[3]{{32768}} = 32$.

Note: Another way to solve this problem is: The given number is 1331. Its unit place is 1, so it’s cube root will also have unit place as 1. The cube root could be 1, 11, 21, 31…. also, keep that in mind, ${10^3} = 1000$ (It’s easy to remember). 1331 is close to 1000 so, we’ll try to calculate${11^3}$ . ${11^3} = 11 \times 11 \times 11 = 1331$ . Bingo! So, 11 is the cube root of a given number.