Question

Find the cube of following numbers using Nikhilam Sutra.
(a) 43 (b) 32 (c) 48

Answer 1

Hint: First of all, we will divide the given number with its nearest base value example 10, 100 etc. Then, from that we will find the value of sub base by using the formula, $s=\dfrac{\text{number}}{\text{base}}$. Then, we will find the difference between original number and the number obtained from the product of base value and sub-base value i.e. $d=\left[ \text{original number}-\left( \text{value of sub-base}\times \text{base} \right) \right]$ and then by substituting the values in formula of cube i.e. $\text{Cube}={{s}^{2}}\left( No.+2d \right)/s\times 3{{d}^{2}}/{{d}^{3}}$, and solving as per Vedic math’s method we will find the answer.

Complete step by step solution:
Before solving the problem, we will understand what Nikhilam Sutra is, so Nikhilam Sutra stipulates subtraction of a number from the nearest power of 101 i.e. 11,101 etc. The powers of 11 from which the difference is calculated is called the Base. If the given number is 104, the nearest power of 11 is 101 and is the base. Hence the difference between the base and the number is 4, which is Positive and it is called NIKHILAM.
Now, the formula of Nikhilam Sutra can be given as,
$\text{Cube}={{s}^{2}}\left( No.+2d \right)/s\times 3{{d}^{2}}/{{d}^{3}}$
Where, s is sub-base and d is difference.
Sub base can be given be given as, ratio of number to that of the base of number which can be seen mathematically as, $s=\dfrac{\text{number}}{\text{base}}$
And difference d, can be given as, difference between original number and the product of value of sub-base and base, which can be given mathematically as,
$d=\left[ \text{original number}-\left( \text{value of sub-base}\times \text{base} \right) \right]$
Now, we will calculate the cube of numbers,
(a) 43
As 43 is a 2-digit number the base is 10 so, sub base can be given as, $s=\dfrac{\text{number}}{\text{base}}$, so on substituting the value we will get,
$s=\dfrac{\text{number}}{\text{base}}=\dfrac{43}{10}=4.3\approx 4$
Now, difference d can be given as, $d=\left[ \text{original number}-\left( \text{value of sub-base}\times \text{base} \right) \right]$, on substituting the values we will get,
$d=\left[ 43-\left( 4\times 10 \right) \right]=\left[ 43-40 \right]=3$
Now, substituting the values in main expression we will get,
$\text{Cube of 43}={{\left( 4 \right)}^{2}}\left( 43+2\times 3 \right)/4\times 3{{\left( 3 \right)}^{2}}/{{3}^{3}}$
$\text{Cube of 43}=16\left( 49 \right)/108/27=784/108/27$
Now, we will consider only last digit of 27 i.e. 7, so 2 will be carry forwarded to the number on the left i.e. 108, so the number will become,
 $\text{Cube of 43}=784/108+2/7$.
Now, again we will consider only the last digit of number 110 i.e. 0, so, 11 will be carry forwarded to the number on the left i.e. 784, so the number will become,
$\text{Cube of 43}=784+11/0/7$
$\Rightarrow \text{Cube of 43}=795/0/7$
As, all the numbers in second and third column are in single digit we can consider this as our final answer which can be written as,
$\Rightarrow \text{Cube of 43}=795/0/7=79507$
 (b) 32
Now, again solving in same way, 32 is a 2-digit number the base is 10 so, sub base can be given as, $s=\dfrac{\text{number}}{\text{base}}$, so on substituting the value we will get,
$s=\dfrac{\text{number}}{\text{base}}=\dfrac{32}{10}=3.2\approx 3$
Now, difference d can be given as, $d=\left[ \text{original number}-\left( \text{value of sub-base}\times \text{base} \right) \right]$, on substituting the values we will get,
$d=\left[ 32-\left( 3\times 10 \right) \right]=\left[ 32-30 \right]=2$
Now, substituting the values in main expression we will get,
$\text{Cube of 32}={{\left( 3 \right)}^{2}}\left( 32+2\times 2 \right)/3\times 3{{\left( 2 \right)}^{2}}/{{2}^{3}}$
$\text{Cube of 32}=9\left( 36 \right)/36/8=324/36/8$
Now, here, the last digit is already a single digit, so we will directly consider the number 36. So, in 36, the last digit is 6, so 3 will be carried forwarded to the number on the left i.e. 324. So, the number will become,
$\text{Cube of 32}=324+3/6/8$
$\Rightarrow \text{Cube of 32}=327/6/8$
As, all the numbers in second and third column are in single digit we can consider this as our final answer which can be written as,
$\Rightarrow \text{Cube of 32}=327/6/8=32768$
(c) 48
Now, again solving in same way, 48 is a 2-digit number the base is 10 so, sub base can be given as, $s=\dfrac{\text{number}}{\text{base}}$, so on substituting the value we will get,
$s=\dfrac{\text{number}}{\text{base}}=\dfrac{48}{10}=4.8\approx 4$
Now, difference d can be given as, $d=\left[ \text{original number}-\left( \text{value of sub-base}\times \text{base} \right) \right]$, on substituting the values we will get,
$d=\left[ 48-\left( 4\times 10 \right) \right]=\left[ 48-40 \right]=8$
Now, substituting the values in main expression we will get,
$\text{Cube of 48}=16\left( 64 \right)/12\left( 64 \right)/{{8}^{3}}$
$\text{Cube of 48=1024}/768/512$
Now, here in number 512, the last digit is 2, so we will carry forward the rest of the number i.e. 51 to the number on the left i.e. 768.
$\text{Cube of 48}=1024/768+51/2$
$\Rightarrow \text{Cube of 48}=1024/819/2$
Now, the last digit in number 819 is 9, so we will carry forward rest of the number to the number on the left i.e. 1024, so the number will become,
$\Rightarrow \text{Cube of 48}=1024+81/9/2$
$\Rightarrow \text{Cube of 48}=1105/9/2$
As, all the numbers in second and third column are in single digit we can consider this as our final answer which can be written as,
$\Rightarrow \text{Cube of 48}=1105/9/2=110592$

Note: Student might miss interpret the formula of finding cube using Vedic math’s i.e. $\text{Cube}={{s}^{2}}\left( No.+2d \right)/s\times 3{{d}^{2}}/{{d}^{3}}$ as, all the terms in division to each other. By considering that and solving it we might get a cube of 48 as $\Rightarrow \text{Cube of 48}=1105/9/2=60.8333$, which is wrong. So, students must follow Vedic math’s method before solving such problems.