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

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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)