
Steps  62^{2}  
1.  Pair the digits, from right to left  in ascending order; and then, from left to right in descending order.  In this case, the pairs are {2},{62},{6} Pairs are obtained by including one digit at a time from right to left (ascending), and then again excluding one digit at a time from left to right (descending) 

2.  Find duplexes for each of the pairs.  In this case, the Duplexes are 4, 24, 36 (Calculating a Duplex is explained below) 

3.  Sum the duplexes by settingup one zero at a time, by simple carry over method.  In this case, 4 + 240 + 3600 = 3844, which is the answer! 
For ndigit number, the Duplex is: 1. For evennumber of digits: Sum of twice the digits, that are equidistant from the ends, multiplied. 2. For oddnumber of digits: Sum of twice the digits, that are equidistant from the ends, multiplied  summed with centraldigit multiplied with itself, that is squared. Then, for a singledigit number (m), Duplex is m^{2} For a doubledigit number (mn), Duplex is 2(m × n) For a 3digit number (mno), Duplex is 2(m × n) + o^{2} For a 4digit number (mnop), Duplex is 2(m × p) + 2(n × o) For a 5digit number (mnopq), Duplex is 2(m × q) + 2(n × p) + o^{2} And so on. As examples, Duplex of 3 is 3^{2} = 9 Duplex of 6 is 6^{2} = 36 Duplex of 23 is 2(2 × 3) = 2 × 6 = 12 Duplex of 64 is 2(6 × 4) = 2 × 24 = 48 Duplex of 128 is 2(1 × 8) + 2^{2} = 16 + 4 = 20 Duplex of 305 is 2(3 × 5) + 0^{2} = 30 + 0 = 30 Duplex of 4231 is 2(4 × 1) + 2(2 × 3) = 8 + 12 = 20 Duplex of 7346 is 2(7 × 6) + 2(3 × 4) = 84 + 24 = 108 Note: That the Duplex of two numbers can be the same (like 128 and 4231 above), especially for similar numbers (like 23 and 32). 
Note that the pairs are {4}, {34}, {234}, {23}, {2} Duplex of 4 is 16 Duplex of 34 is 24, setup one zero to 240 Duplex of 234 is 16 + 9 = 25, setup two zeroes to 2500 Duplex of 23 is 12, setup three zeroes to 12,000 Duplex of 2 is 4, setup four zeroes to 4,0000 Sum of Duplexes = 16 240 2500 12000 + 40000  54756 Thus, 234^{2} = 54,756 
Note that the pairs are {6}, {26}, {426}, {1426}, {142}, {14}, {1} Duplex of 6 is 36 Duplex of 26 is 24, setup one zero to 240 Duplex of 426 is 48 + 4 = 52, setup two zeroes to 5200 Duplex of 1426 is 12 + 16 = 28, setup three zeroes to 28,000 Duplex of 142 is 4 + 16 = 20, setup four zeroes to 20,0000 Duplex of 14 is 8, setup five zeroes to 8,00000 Duplex of 1 is 1, setup six zeroes to 1,000000 Sum of Duplexes = 36 240 5200 28000 200000 800000 + 1000000  2033476 Thus, 1426^{2} = 2,033,476 Note that, the professional approach is to write the answer from right to left, without adding the Zeroes  summing the Duplexes on the go: Step 1: 36; 6 (carry 3) Step 2: 3 + 24 = 27; 7 (carry 2) Step 3: 2 + 52 = 54; 4 (carry 5) Step 4: 5 + 28 = 33; 3 (carry 3) Step 5: 3 + 20 = 23; 3 (carry 2) Step 6: 2 + 8 = 10; 0 (carry 1) Step 7: 1 + 1 = 2 Hence, 2033476 
Assuming a twodigit number, mn, then N = 10m + n As in example above, 62 = 60 + 2 Now, N^{2} = (10m + n)^{2} = 100m^{2} + 2(10mn) + n^{2} = 100(m^{2}) + 10(2mn) + n^{2} This is exactly what this Upasūtra makes us do: 62^{2} = (100 × 36) + (10 × 24) + 4 = 3844 Assuming a 3digit number, mno, then N = 100m + 10n + o As in example above, 234 = 200 + 30 + 4 Now, N^{2} = (100m + 10n + o)^{2} = ((100m + 10n) + o)^{2} = (100m + 10n)^{2} + 2((100m + 10n) × o) + o^{2} = [10000m^{2} + 2(100m × 10n) + 100n^{2}] + [2(100mo) + 2(10no)] + o^{2} = [10000(m^{2}) + 1000(2mn) + 100(n^{2})] + [100(2mo) + 10(2no)] + o^{2} = 10000(m^{2}) + 1000(2mn) + 100((2mo) + n^{2}) + 10(2no) + o^{2} This is exactly what this Upasūtra makes us do: 234^{2} = (10000 × 4) + (1000 × 12) + (100 × 25) + (10 × 24) + 16 = 54756 Note that, as the number of digits increase  so does the number of steps that this Upasūtra makes us perform  in the same pattern as above. 

Steps  14^{3}  
1.  Find the ratio of the two digits.  In this case, the Ratio is 1:4  
2.  Pair 4 terms, the first being the cube of 1st digit, and the remaining terms as Geometric Progression as the Ratio.  In this case, {1},{4},{16},{64} As above, 1^{3} = 1, and 1 × 4 = 4; 4 × 4 = 16; 16 × 4 = 64 

3.  Pair 4 more terms, with 2nd and 3rd terms being the double of the original values. The others being setup to zero.  In this case, it is {0},{8},{32},{0}  
4.  Sum the terms, by simple carry over method.  In this case, {1 + 0},{4 + 8},{16 + 32},{64 + 0} = 2744, which is the answer! Note: [1,4,16,64] + [0,8,32, 0]  Step 1: 64 + 0 = 64; 4 (carry 6) Step 2: 6 + 16 + 32 = 54; 4 (carry 5) Step 3: 5 + 4 + 8 = 17; 7 (carry 1) Step 4: 1 + 1 + 0 = 2 Hence, 2744 
Clearly, the ratio is 1:8 And, 1^{3} = 1 So, the first pair of terms are: {1},{8},{64},{512} And, the second pair of terms are: {0},{16},{128},{0} Sum of the pairs is: 5832 Note: [1, 8, 64,512] + [0,16,128, 0]  Step 1: 512 + 0 = 512; 2 (carry 51) Step 2: 51 + 64 + 128 = 243; 3 (carry 24) Step 3: 24 + 8 + 16 = 48; 8 (carry 4) Step 4: 4 + 1 + 0 = 5 Hence, 5832 Thus, 18^{3} = 5,832 
Clearly, the ratio is 2:8 = 1:4 And, 2^{3} = 8 So, the first pair of terms are: {8},{32},{128},{512} And, the second pair of terms are: {0},{64},{256},{0} Sum of the pairs is: 21952 Note: [8,32,128,512] + [0,64,256, 0]  Step 1: 512 + 0 = 512; 2 (carry 51) Step 2: 51 + 128 + 256 = 435; 5 (carry 43) Step 3: 43 + 32 + 64 = 139; 9 (carry 13) Step 4: 13 + 8 + 0 = 21 Hence, 21952 Thus, 28^{3} = 21,952 
Clearly, the ratio is 7:7 = 1:1 And, 7^{3} = 343 So, the first pair of terms are: {343},{343},{343},{343} And, the second pair of terms are: {0},{686},{686},{0} Sum of the pairs is: 456533 Note: [343,343,343,343] + [ 0,686,686, 0]  Step 1: 343 + 0 = 343; 3 (carry 34) Step 2: 34 + 343 + 686 = 1063; 3 (carry 106) Step 3: 106 + 343 + 686 = 1135; 5 (carry 113) Step 4: 113 + 343 + 0 = 456 Hence, 456533 Thus, 77^{3} = 456,533 

If we split the number into two, say: 1 and 21 Clearly, the ratio is 1:21 And, 1^{3} = 1 So, the first pair of terms are: {1},{21},{441},{9261} And, the second pair of terms are: {0},{42},{882},{0} Sum of the pairs is: 1771561 Note: [1,21, 441,9261] + [0,42, 882, 0]  [1,63,1323,9261] Note again, that since the second part has 2 digits, the working Base is 100 So, [1,63,1323,9261] = 1,000000 + 63,0000 + 1323,00 + 9261 = 1771561 Thus, 121^{3} = 1,771,561 
If we split the number into two, say: 2 and 78 Clearly, the ratio is 1:39 And, 2^{3} = 8 So, the first pair of terms are: {8},{312},{12168},{474552} And, the second pair of terms are: {0},{624},{24336},{0} Sum of the pairs is: 21484952 Note: [8,312,12168,474552] + [0,624,24336, 0]  [8,936,36504,474552] Note again, that since the second part has 2 digits, the working Base is 100 So, [8,936,36504,474552] = 8,000000 + 936,0000 + 36504,00 + 474552 = 21484952 Thus, 278^{3} = 21,484,952 
If we split the number into two, say: 11 and 48 Clearly, the ratio is 11:48 = 1:(48/11) And, 11^{3} = 1331 So, the first pair of terms are: {1331},{5808},{25344},{110592} And, the second pair of terms are: {0},{11616},{76032},{0} Sum of the pairs is: 1512953792 Note: [1331, 5808,25344,110592] + [ 0,11616,50688, 0]  [1331,17424,76032,110592] Note again, that since the second part has 2 digits, the working Base is 100 So, [1331,17424,76032,110592] = 1331,000000 + 17424,0000 + 76032,00 + 110592 = 1512953792 Thus, 1148^{3} = 1,512,953,792 Again, if we split the number into two, say: 1 and 148 Clearly, the ratio is 1:148 And, 1^{3} = 1 So, the first pair of terms are: {1},{148},{21904},{3241792} And, the second pair of terms are: {0},{296},{43808},{0} Sum of the pairs is: 1512953792 Note: [1,148,21904,3241792] + [0,296,43808, 0]  [1,444,65712,3241792] Note again, that since the second part has 3 digits, the working Base is 1000 So, [1,444,65712,3241792] = 1,000000000 + 444,000000 + 65712,000 + 3241792 = 1512953792 Thus, 1148^{3} = 1,512,953,792 
Assuming a twodigit number, mn, then N = 10m + n As in examples, 14 = 10 + 4, 18 = 10 + 8, 28 = 20 + 8, 77 = 70 + 7 Now, N^{3} = (10m + n)^{3} = (10m + n)(10m + n)^{2} = (10m + n)(100m^{2} + 2(10mn) + n^{2}) = [1000m^{3} + 2(100m^{2}n) + 10mn^{2}] + [100m^{2}n + 20mn^{2} + n^{3}] = 1000m^{3} + 3(100m^{2}n) + 3(10mn^{2}) + n^{3} This can be written as: = 1000m^{3} + 100m^{2}n + 10mn^{2} + n^{3} + 100(2m^{2}n) + 10(2mn^{2}) This is exactly what this Upasūtra makes us do: 14^{3} = 1000(1) + 100(4) + 10(16) + 64 + 100(8) + 10(32) The others, also being 2digit numbers, are similar  as above. Note that, the ratios are important because that is how the power of n is being distributed. But more importantly, m^{3}:m^{2}n = m^{2}n:n^{3} = m:n Similarly, numbers with more digits may be proven. 
« Upasūtra 15 