Consecutive Cube Conversion
If three consecutive integers are multiplied together and the middle number is added, will it always produce a cube?
For example, the first three consecutive positive integers satisify the condition: ( 1 × 2 × 3 ) + 2 = 8 = 2 3 .
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5 solutions
The answer is Yes
Considering three consecutive integers n , n + 1 and n + 2 we get. n ( n + 1 ) ( n + 2 ) + ( n + 1 ) = ( n + 1 ) ( n ( n + 2 ) + 1 ) = ( n + 1 ) ( n 2 + 2 n + 1 ) = ( n + 1 ) ( n + 1 ) 2 = ( n + 1 ) 3
Much easier to consider the integers n − 1 , n , n + 1 . Think it's been done already somewhere...
Let middle term be x therefore 3 nos. are x-1,x & x+1
given
x+ (x)(x-1)(x+1) =(x^2-1)x+x =x^3-x+x =x^3 HENCE PROVED
Let n equal the middle term
The product of three consecutive digits can be represented as:
( n − 1 ) n ( n + 1 ) = ( n 2 − 1 ) n = n 3 − n
Then we add the middle term n
n 3 − n + n = n 3
Is it a cube?
3 n 3 = n
n is a whole number and the cube root of the equation hence when three consecutive integers are multiplied together and the middle integer is added; it will always produce a cube number.
n(n+1)(n+2) + (n+1) = (n+1)^3
A quicker method is to view them in the form of ( x − 1 ) x ( x + 1 ) + x = ( x 3 − x ) + x = x 3