# using ramanujan hardy method find the cubes ofa) 13832b) 4104

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:

$⇒$( 23 + 2433  = ( 183 + 203 )3

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

$⇒$[ (23)3 + (243)3 + 3$×$2$×$243 ( 23 + 243 ) ]  = [ (183)3 + ( 203)3 +3 $×$183 $×$203( 183 + 203 )]

$⇒$[ 29 + 249 + 3$×$2$×$243 ( 23 + 243 ) ]  = [ 189 + 209+3 $×$183 $×$203( 183 + 203 )]

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

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

$⇒$   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

$⇒$[ (23)3 + (163)3 + 3$×$2$×$163 ( 23 + 163 ) ]  = [ (93)3 + ( 153)3 +3 $×$93 $×$153( 93 + 153 )]

$⇒$[ 29 + 169 + 3$×$2$×$163 ( 23 + 163 ) ]  = [ 99 + 159 +3 $×$93 $×$153( 93 + 153 )]

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

$⇒$[ 512 + 68719476736 + 403439616 ] = [ 387420489 + 38443359375 + 30292137000 ]

$⇒$69122916864 = 69122916864

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

• 0

Can I do it by prime factorisation.

• -3

No

• -2
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