Can you prove?
Consider any three consecutive natural numbers, then cube of the largest number is the sum of the cubes of the other two numbers.
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2 solutions
It is always FALSE :
Let three consecutive number be a , b , c and a < b < c so :
c 3 = a 3 + b 3
But it have no solution because of Fermat Last Theorm.
Ok, If I take numbers that is
1
,
2
,
3
, here largest of these is
3
Now, According to question,
3
3
=
(
1
+
2
)
3
2
7
=
2
7
????
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You must understand the difference between sum of the cube and cube of the sum . Let the two number be x and y
- sum of the cube = x 3 + y 3
- cube of the sum = ( x + y ) 3
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Oh! Thank you for correcting me here.I always have doubt in sum of cubes and cubes of sum.I will remember this now.
Without using Fermat's Last Theorem, let the number be ( x − 1 , x , x + 1 ) . We will get ( x − 1 ) 3 + x 3 = ( x + 1 ) 3 , or simply x 3 = 6 x 2 + 2 . The real x being x = 2 + 3 9 + 1 7 + 3 9 − 1 7 , not being natural number. It is then proved that no natural number x exists.