Prove: The difference of the cubes of two successive integers is odd.
(n + 1)3-n3 = [ ? ]n2 + [ ]n + 1
= [ ](n + 1) + 1
= 6m + 1 (since n or n+1 is even)
= odd

(n + 1)³ - n³ = (n³ + 3n² + 3n + 1) - n³

… = 3n² + 3n + 1

… = 3n (n + 1) + 1

… = 6m + 1

where m is another integer. We can say this because, as the second-to-last step says, either n or n + 1 is even, which makes their product also even, so we can write it as a multiple of 2:

n (n + 1) = 2m

Then 6m is even, and adding 1 to it makes it odd.

picture below

answer: the answer is simply nothing.

step-by-step explanation: because there is no ending to this question

gimme brainliest > : 3