If $(a-1),a,(a+1)$ are consecutive positive integers, $ (a+1)^3 \neq a^3 + (a-1)^3$ I had to prove the following statement: 

If $(a-1),a,(a+1)$ are consecutive positive integers, $(a+1)^3 \neq a^3 + (a-1)^3$

My attempt at the solution was to first expand each side to get 
$$a^3 + 3a^2 + 3a + 1 \neq 2a^3 - 3a^2 +3a - 1\\
      0 \neq a^3 - 6a^2 - 2$$
However, $a^3 - 6a^2 - 2$ does hit the $x$-axis at $a = 6.0546$.

Does that mean that the statement is incorrect?


All you need to do is show that the equation $a^3-6a^2-2=0$ has no positive integer solution.  If it did, then you'd have $a^2(a-6)=2$.  But $a^2\mid2$ implies $a=1$, in which case $a^2(a-6)=-5\not=2$.
