Vedic Math - Cube Roots of more than 6-Digit Number - Part III

Note: Vedic Math Blog has been moved to Please bookmark the new address for new and existing blogs.

In this article, we are going to learn an interesting Mathematical technique to find, if the given number is a perfect cube or not. It is very important step while computing cube roots. Infact, before applying any method to find the cube root, we have to check whether it is perfect cube or not and then accordingly we choose the technique. For example, following scenario tells us the importance of finding perfect cube step while computing the cube root.

Example : 1728 has cube root 12 since two groups are 1 and 728. From 728, we derive last digit as                 2 from 1 (first group), we derive first digit as 1.
               So, cube root of 1728 is 12.
 But now, if number is 1278, which again has two groups: 1 and 278. It can derive the same last digit as 2 and first digit as 1 , which implies that cube root of 1278 is 12, which is not true because technique stands true for perfect cube root only.

There is a simple technique to check whether the number is perfect cube or not. For this, we add the digits of the number. See the below chart in which we add the digits of cubes from 1 to 10.

Above example shows that sum of digits of a perfect cube is either 1, 8 or 9. However, it is not true that all numbers which sum to 1,8 or 9, will be perfect cube.

For example,
Sum of digits of 1728 and 1278 are same i.e.(1+7+2+8) = (18) = 9 . But 1278 is not a perfect cube.

Hence if sum of digits of a number is not 1,8 or 9, we are very sure that the number is not a perfect cube. However, a number may not be perfect cube root even if sum of digits is 1,8 or 9. To scrutinize that, we need to apply factorisation. If number is small like 1278, factorisation is good method. See below:

For bigger numbers, factorisation could be time consuming technique. Hence, for large numbers, we shall apply general method of finding the cube of root.

Case 2 : Cube root for all the cubes, whether perfect cubes or not.    (Case 1 discussed in last two articles)
From last two articles, we conclude about the sequence of digits (a+b+c)³ as:
(1) The first place by a³
(2) The second place by 3a2b
(3) The third place by 3ab2+3a2c
(4) The fourth place by 6abc+b³
(5) The fifth place by 3ac2+3b2c
(6) The sixth place by 3bc2
(7) The seventh place by c³ ; and so on.

In 'General Technique', we find Dividends(D), Quotients(Q), and Remainders(R). Steps involved as:
(1) First determine D, Q and R
(2) From the second dividend, no deduction is to be made.
(3) From the third, subtract 3ab2
(4) From the fourth, deduct 6 abc+b³
(5) from the fifth, subtract 3ac2+3b2c
(6) from the sixth, deduct 3bc2
(7) from the seventh, subtract c³. ; and so on.

(a) Quotient(Q) is closest minimum exact cube to the first cube i.e. 'F' term used in last two articles.
(b) And, Reminder(R) is the difference between the first group and closest minimum exact cube.
(c) Dividend(D) is found by multiplying the 'Square of Quotient(Q)' by 3 (Q2*3)

Vedic Math - Cube Roots of more than 6-Digit Number - Part II

Note: Vedic Math Blog has been moved to Please bookmark the new address for new and existing blogs.

In last article, we have discussed the method to find cube root of more than 6-digit numbers; especially the odd numbers. Today we shall discuss the procedure for even numbers. In this procedure, only two extra steps are added, one in the beginning and other at the end. Rest all is same.

Procedure: As first added step, we keep on dividing the number by 8 till we get an odd cube. Following it, same method of successive elimination of the digits will apply. At the end, multiply the cube root by 8 to obtain the cube root of the original number.

Example :  2840362499528
First, we continue without using those two additional steps, which will help you to understand the problems arises while dealing with even cubes.

The cube root of the cube 2,840,362,499,528      (say, F + J + H + M + L )
Here,  N=5       (means that cube root will be of 5 digits number)
L=2                  (i.e. 2³=8, matching with the last digit of the last group '528')
and F=1            (i.e, 1³=1, nearest cube of first group '1')

Step1 : L=2 & L³=8. Subtracting this,              
Step2 : 3L2M=12M (substituting L = 2)
            Hence, 12M = Number ending with 2
    Here M is either '1' or '6'      (ambiguous values)
            Lets take 6  (pure gamble)
    Now, Deducting 3L2M = 12M = 72  
Step3 : 3LM2+3L2H = 12H + 216 (substituting L = 2, M = 6)
            12H + 216 = Number ending with 8
      12H = Number ending with 2                                    
            Here H is either '1' or '6'                                      
    Lets take 1 (again gamble)

    Now, Subtract 3LM2+3L2H = 12H + 216
                                                         = 228
Step4 : 3L2J+6LMH+M³ = 12J+12+216          
                                             = 12J+228
        12J+228 = Number ending with 6
12J = Number ending with 8
        Here J is either '4' or '9'
        Lets take J = 4  

Since we already know 'F' , so no need to know the expansion of (F+J+H+M+L)³ 
Therefore, cube root is 14162        (F=1, J=4, H=1, M=6, L=2)