Consecutive cubes!
n 3 + ( n + 1 ) 3 + ( n + 2 ) 3 = ( n + 3 ) 3
How many real solution(s) are there to the above equation?
The answer is 1.
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1 solution
From which we get 3 3 + 4 3 + 5 3 = 6 3 . Nice solution. +1
How do you get from 2 ( n 3 − 6 n − 9 ) = 0 to 2 ( n − 3 ) ( n 2 + 3 n + 3 ) = 0 ?
n 3 + ( n + 1 ) 3 + ( n + 2 ) 3 = ( n + 3 ) 3
n 3 + ( n 3 + 3 n 2 + 3 n + 1 ) + ( n 3 + 6 n 2 + 1 2 n + 8 ) = ( n 3 + 9 n 2 + 2 7 n + 2 7 )
⟹ 3 n 3 + 9 n 2 + 1 5 n + 9 = n 3 + 9 n 2 + 2 7 n + 2 7
⟹ 2 n 3 − 1 2 n − 1 8 = 0 ⟹ 2 ( n − 3 ) ( n 2 + 3 n + 3 ) = 0
The solutions for n are 3 , 2 3 + 3 i , 2 3 − 3 i . Thus, there is only 1 real solution.