# Find all primes of the form n^3-1 for natural number'n'

Please find below the solution to the asked query:

We know : a

^{3}- b

^{3}= ( a - b ) ( a

^{2}+ ab + b

^{2})

So,

( n

^{3}- 1 ) = ( n

^{3}- 1

^{3}) = ( n - 1 ) ( n

^{2}+ n + 1 )

So, We get ( n

^{3}- 1 ) as prime number only if one of ( n - 1 ) and ( n

^{2}+ n + 1 ) is equal to 1 , So

At n

^{2}+ n + 1 = 1 ,

n

^{2}+ n = 0 ,

n ( n + 1 ) = 0 , So

n = 0 and n = - 1 , Both are not a natural numbers ,

And At n - 1 = 1 ,

n = 2 and at n = 2 , That is a natural number we get

( n

^{3}- 1 ) = 2

^{3}- 1 = 8 - 1 = 7 , That is a prime number

So,

We get only one prime number that is in form of n

^{3}- 1 where n is a natural number .

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