Applications of Prime factor method

Sum of two cubes   can be made to equal with another sum of two cubes, i.e., some numbers can be expressed as sum of two cubes in two different ways.After exposing the personality of 1729, these numbers are generally called as Ramanujan Numbers.
If the Ramanujan number is the product of three different  odd prime numbers, it will be odd.
a3 + b3 = c3 + d3  = R = Po1 Po2 Po3
(a+b) (a2 - ab - b2)  =(c+d) (c2 - cd + d2) = Po1 Po2 Po3
Since the prime factors cannot be factorized further,  (a+b) and (c+d) must be equal to  one of the three prime factors independently
If a+b= Po3
a2 - ab +b2  = Po1 Po2
Substituting the value of a or b  obtained from the first equation in the second equation,
3b2 - 3bPo1 + Po32 - Po1 Po2   = 0
Solving for b,
b = [3Po3 + √ 12Po1 Po2   - 3Po32 ]/6
a =  [3Po3 - √ 12Po1 Po2   - 3Po32 ]/6
a and b to be whole number, the term inside the square root must be a square number
If c + d= Po2
c2 - cd + d2  = Po1 Po3
It gives c = [3 Po2 +√ 12Po1 Po3   - 3Po22 ]/6 and d =[3 Po2 -√ 12Po1 Po3   - 3Po22 ]/6

  7  x   13 x   19    =        1729    =   13 + 123    =     93  +   103
 13 x   37 x   43    =      20683    = 103 + 273    =   193  +   243
 73 x 157 x 211    =  2418271    = 233 +1343   =   953  + 1163  
103x 241 x 307    =  7620661    = 453 +1963   =  1333 + 1743

When the odd Ramanujan number is expressed as the product of two prime numbers and one square of a prime number or a product of two prime numbers

19x 72 x 43   = 40033 = 163 + 333 = 93 + 344
132 x 67 x 163 = 1845649 = 423 + 1213 = 493 + 1203
(7x19) x 379 x 409 = 20616463 = 1113 + 2683 = 1513 + 2583

In the case of even Ramanujan numbers, 2n  will be one of the factors.If the other two facgtors are prime, then, (a+b) or (c+d) will be even . It can be expressed by the entire 2n  and it may include or not one of the prime factors.

 13 x 79 x 26 = 65728 = 123  + 403 = 313 + 333