*Cubing by Yaavadunam*

This method is
similar to squaring by *Yaavadunam*. It is just modified a bit,
as we shall see in the next few examples.

Consider 13^{3}

Step 1 : Consider nearest base (here 10).

Step 2 : As 13 has a excess of '3' (13 - 10 = 3), we double the excess and add the original number (13) to it, and put it on the LHS.

Therefore we get 13 + 6 = 19

Step 3 : Now find the new excess. In this case it is 19-10 = 9. Now multiply this with the original excess to get the middle part of the answer.

Therefore we get 9 * 3 = 27

Step 4 : Now cube the original excess and put it as the last part

*Carry over any big numbers and total to get the answer.*

19 7 72 221 9 7

Therefore 13^{3}
= 2197

Now consider 47^{3}

As in 'Nikhilam' and Squaring, we use 'Aanurupyena' here.

1) Let the
* main base* be 10 and the * working base* be 50

therefore the ratio

x = (Main Base)/(Working Base) = 10/50 = 1/5

2) Excess is -3 (47 - 50 = -3). Double the excess and add the original number (here 47) to it.

We get 47 - 6 = 41.

The Base correction for this part is achieved by dividing by x^{2}
.

therefore we get 41/(1/25) = 41 * 25 = 1025

3) Excess in the new uncorrected number (41 - 50 = -9) is multiplied by the original excess(-3) to obtain the second part.

Therefore we get -9 * -3 = 27

The Base correction for this part is achieved by dividing by x .

therefore we get 27 * 5 = 135

4) The third part is obtained by cubing the excess.

(-3)^{3} =
-27

5) Carry over the extra numbers and total to obtain the final answer

1025 0 0 13 5 0-2 71038 2 3

Therefore the final answer is 103823