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. 

Now, you just have to show that the cubic $x^3-6x^2-2 = 0$ has no positive integer roots. 
By the rational root theorem, the only possible rational roots are $x = \pm 1, \pm 2$. Hence, the only possible positive integer roots are $x = 1,2$. Is $x = 1$ or $x = 2$ a root? If not, then there are no positive integer solutions $x = n$ to $x^3-6x^2-2 = 0$.
Alternatively, you can use Fermat's Last Theorem, but that is overkill.
