using ramanujan hardy method find the cubes of

a) 13832

b) 4104

Answer:
According to Ramanujan Hardy method , some special numbers can be expressed as sum of two positive cubes in two distinct ways:
 
a )  13832
We can expressed it by Ramanujan Hardy method As:

13832 = 2³ + 24³  = 18³ + 20³

Now to find cube of 13832 , taking whole cube of both side As:

$\Rightarrow$( 2³ + 24³ ) ³  = ( 18³ + 20³ )³

As we know ( a + b )³ = a³ + b³ + 3ab(a + b ) , so we get

$\Rightarrow$[ (2³)³ + (24³)³ + 3$\times$2³ $\times$24³ ( 2³ + 24³ ) ]  = [ (18³)³ + ( 20³)³ +3 $\times$18³ $\times$20³( 18³ + 20³ )]

$\Rightarrow$[ 2⁹ + 24⁹ + 3$\times$2³ $\times$24³ ( 2³ + 24³ ) ]  = [ 18⁹ + 20⁹+3 $\times$18³ $\times$20³( 18³ + 20³ )]

$\Rightarrow$[  512  + ​2641807540224 + ​331776 ( 8 + ​13824 )] = [ ​198359290368 + ​512000000000 + ​139968000 ( ​5832 + ​8000)]

$\Rightarrow$[ 512 + ​2641807540224 + ​4589125632 ]  = [ ​198359290368 + ​512000000000 + ​1936037376000

$\Rightarrow$   2646396666368   = 2646396666368

So,
( 13832 )³ = 2646396666368                  ( Ans )


b ) 4104
We can expressed it by Ramanujan Hardy method As:

4104 = 2³ + 16³ = 9³ + 15³

Now to find cube of 4104 , taking whole cube of both side As:

( 2³ + 16³ ) = ( 9³ + 15³ )

As we know ( a + b )³ = a³ + b³ + 3ab(a + b ), So we get

$\Rightarrow$[ (2³)³ + (16³)³ + 3$\times$2³ $\times$16³ ( 2³ + 16³ ) ]  = [ (9³)³ + ( 15³)³ +3 $\times$9³ $\times$15³( 9³ + 15³ )]

$\Rightarrow$[ 2⁹ + 16⁹ + 3$\times$2³ $\times$16³ ( 2³ + 16³ ) ]  = [ 9⁹ + 15⁹ +3 $\times$9³ $\times$15³( 9³ + 15³ )]

$\Rightarrow$[ 512 + ​68719476736 + ​98304 ( 8 + ​4096 ) ] = [ ​387420489 + ​38443359375 + ​7381125( 729 + ​3375 ) ]

$\Rightarrow$[ 512 + 68719476736 + 403439616 ] = [ 387420489 + 38443359375 + 30292137000 ]

$\Rightarrow$69122916864 = 69122916864

So,
( 4104 )³ = ​69122916864                ( Ans )