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

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Back after a long break. In previous article, we learn how to find the cube root of 4 or 5 or 6 digits perfect cubes. Let us continue it further and discuss how to find the cube root of perfect or imperfect cubes.

In this article, we shall learn to find the cube roots for:
1. Cube root of perfect cubes, for any number of digits.
2. Cube root for all the cubes, whether perfect cubes or not.

To summarize what we have learned till now for cube root:

Arrange the given number in three-digit groups, starting from right to left. A single digit, if any left over at the left hand side, is counted as a simple group itself. The number of  digits in the cube root will be the same as the number of digit-groups  in the given  number itself.

  • 169 will count as 1 group 
  • 1 258 will count as 2 groups 
  • 43 781 will count as 2 groups
  • 2 154 890 will count as 3 groups

If the given number has 'n' digits, its cube root will be having n/3 or (n+1)/3 digits. Also remember few other points from previous article:

The Cubes  of the first nine natural numbers    
1³ = 1        2³ = 8        3³ =27        4³ = 64        5³ = 125        6³ = 216        7³ = 343        8³ = 512         9³ = 729        10³ = 1000
From it, we understand that

  • 1,4,5,6,9,0 numbers repeat themselves in the ending of their cubes
  • 2,3,7 and 8 have their complements from 10, at the end of their cube

Let us start with actual technique now. Any number can be written in an algebraic expression. For example, if arithmetical number is 'dcba', it can be written in algebraic form as:
Algebraic Expression is: a + 10b + l00c + 1000d.

Now if we need to find the cube of a number 'cba', algebraically we can expand it like  (a+10b+102c)3. Let us expand it:
(a+10b+102c) 3 =  a3 + 10 (3a2b) + 102 (3ab2+3a2c) + 103 (b3+6abc) + 104 (3ac2+3b2c) + 105 (3bc2) + 106 (c3)

Now removing the powers of ten and putting the result in algebraic form, it tells us the formation of cube as:
(1) The units' place is determined by a³.
(2) The tens' place is contributed by 3 a2b
(3) The hundreds' place is contributed to by 3ab2 + 3a2c
(4) The thousands' place is formed by b³ + 6abc
(5) The ten thousands' place is given by 3ac2 + 3b2c
(6) The hundred thousands' (lakhs') place is constituted of 3bc2 ; and
(7) The millions' place is formed by c³.

The number of zeroes in the various coefficients of the expanded Algebraic Expression are the basis of the formula / analysis.

Case1 :  Cube root of perfect cubes for any number of digits

Suppose we have a cube number n of any number of digits. To find its cube root, find following:
- The number of groups (N) in cube (as we discussed above to make the sets of 3 digits)
- First digit of cube root denoted as 'F' (Nearest cube root of first group from left)
- Last digit of cube root denoted as 'L' (Cube root of last group from left)
- Middle numbers of cube root(i.e. 'M' or 'H' or 'J'....), we shall find using the procedure.

Following are the steps for the procedure:
(i) From the units' place of given number, subtract the L³ (i.e. a³, refer to algebraic expression above); and that eliminates the last digit of the number.
(ii) From the ten's place, we subtract 3L2M (i.e. 3a2b) and thus eliminate the second last digit (penultimate digit).
(iii) From the hundreds' place, we subtract 3LM2 + 3L2F (i.e. 3ab2 + 3a2c) and hence eliminate the pre-penultimate digit.
and so on

Let us understand it with an example.

Example 1 :  83453453

First arrange the numbers in groups i.e. 83,453,453
Here the number of 3 digit groups i.e. N = 3 (that also means that cube root will be of 3 digits number)
Last digit of the number is 3. The last digit of cube root of the number ending with 3 should be 7 (i.e.7³ = 343) So L = 7
The first group is 83 , the closest minimum exact cube to 83 is 64 which is nothing but 4³, therefore, F = 4

Hence N = 3, L = 7, and F = 4

And let us proceed to find the middle number:

Step1 : L=7 so L³=343                                                      
        Now subtract '343' from the last digit

       This eliminates the last digit of number, and give way to calculation of middle numbers

Step2 :   3L2M
= 147M        (substituting L = 7)
Now the last digit of 147M is '1' (which is the last digit of the number '8345311' after subtracting in first step).
            Hence,  147M = Number ending with 1
So M should be 3 ( because 147*3 can give a number ending with 1)
     
Deducting 3L2M = 147M = 147*3 = 441  
     
This eliminates the second last digit of number, and give way to calculation of further numbers.

Step3 :  3LM+ 3L2F = 147F + 189  =  ending with 7 (last digit of '834487' )
             147F  =  Number ending with 8   (i.e.834487-189)
        Hence, F = 4     (147*4 can give a number ending with 8)

 Hence, Cube Root = 437

Note:- The last step can be skipped safely, as we were already aware about the the first digit of cube root i.e. F.

Example 2: 13824
Arrange the number in groups i.e. 13, 824
Here N=2 (which means that cube root will be of 2 digits number)
L = 4 (i.e. 4³ = 64) and
F = 2 (exact cube near to 13 is 8 i.e.2³)
Therefore,  cube root = 24

Example 3 : 76928302277
Arrange the number in groups i.e. 76, 928, 302, 277  (say, F + H + M + L )
Here N = 4 (which means that cube root will be of 4 digits number, FHML)
L = 3 (i.e. 3³ = 27, matching with the last digit of the last group '277')
and F = 4 (i.e, 4³ = 64, nearest cube of first group '76')

Step1 : L=3 & L³=27. Subtracting this,    
Step2 : 3L2M=27M (substituting L = 3)
        Now the last digit of 27M is '5' (which is the last digit of the number '7692830225' after subtracting in first step).
        Hence, 27M = Number ending with 5
So M should be 5 ( because 27*5 can give a number ending with 5)      
     
Now, Deducting 3L2M = 27M                                    
                                   = 135         (27*5)        
Step3 : 3LM2+3L2H = 27H + 225 (substituting L = 3, M = 5)
            27H+225 = Number ending with 9
27H = Number ending with 4   (i.e.769283009-225)  
           Hence,  H = 2   (27*2 can give a number ending with 4)                          
So,  3LM2+3L2H = 27H+225 = 27*2+225 = 279            
Therefore, cube root is 4253    (F=4,  H=2,  M=5,  L=3)            

Note: According to the procedure, we can determine the first digit by the same method of successive elimination of the digits. For this, we have to expand (F + H + M + L)³. And, on expanding, we find that in the 4th step i.e. thousand's place number, 'F' appears to be '3L2F + 6LMH + M³'. We need to deduct this value to eliminate the fourth digit. So,

        3L2F + 6LMH + M³ = 27F+180+125          
                                         = 27F+305
        27F + 305 = Number ending with 3
27F = Number ending with 7
        Hence,   F = 4
        Therefore,  Cube Root = 4253  

[For reference, (103d+102c+10b+a) 3 = a3 + 10 (3a2b) + 102 (3ab2+3a2c) + 103 (b3+6abc+3a2d) + 104 (3ac2+3b2c+6abd) + 105 (3bc2+3b2d+6acd) + 106 (c3+3ad2+6bcd) + 107 (3bd2+3c2d) + 108 (3cd2) + 109 (d3)]           

Note:- Again, in case of perfect cubes, the last step can be skipped safely, as we are already aware about the the first digit of cube root i.e. F.

Important point to notice is that above method can be applied for odd cubes only. If the cube is even, then we would found ambiguous values at each step; like we may get two possibilities of M, H,.... We shall discuss the procedure to find cube root of such even cubes in next article. Till then, practice above procedure and keep posting your queries and share any findings.

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