Prove by contradiction that $(n+1)^3 \not= n^3 +(n-1)^3$ for $3$ consecutive positive integers Prove by contradiction that if $n-1$, $n$, $n+1$ are consecutive positive integers, then the cube of the largest cannot be equal to the sum of the cubes of the other two.
Assume that:
$$ (n+1)^3 = (n-1)^3+n^3 $$
$$n^3+3n^2+3n+1=n^3-3n^2+3n-1+n^3$$
$$3n^2+1=-3n^2-1+n^3$$
$$n^3-6n^2-2=0$$
I don't know how to move from here. Instead of this solution considered that two of the integers are either odd or even but this idea didn't help much. 

$n^3-6n^2-2=0 \implies n^2(n-6)=2$


*

*$n<6 \implies n^2(n-6)<0$

*$n=6 \implies n^2(n-6)=0$

*$n>6 \implies n^2(n-6)\geq49$


Therefore, $\forall{n}\in\mathbb{N}:n^2(n-6)\neq2$
