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?

 A: Suppose $n^3-6n^2 = 2$ for some integer $n$. It is easy to check that $n$ cannot be odd. Hence $n = 2k$ for some integer $k$. We have $8k^3-24k^2 = 2$, or equivalently, $k^3-3k^2 = \frac{1}{4}$, which is a contradiction since $k$ was supposed to be an integer.
A: 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 $a^2(a-6)=-5\not=2$.
A: Perhaps not the shortest or the greatest argument, but it may be helpful to the OP. Let $f(a) = a^3-6a^2-2$. We take the derivative to obtain $f'(a) = 3a^2-12a$. The roots of the derivative are $0$ and $4$. Looking at the sign of the derivative, we observe that $f$ is increasing on $(-\infty,0)$, decreasing on $[0,4)$, and again increasing on $[4,\infty)$. Since $f(0) = -2$, and $f$ is increasing on $(-\infty,0)$, we know there cannot be any roots that are less or equal to $0$. Now, we also know $f$ is decreasing on $[0,4]$ so there can be roots in this interval either. Now we know that the only root is (if it exists) greater than or equal to $4$. Again the fact that $f$ is decreasing on $[0,4]$ shows that $f(4)<0$ (without calculation). Since $f$ grows to infinity as its argument grows to infinity, there is thus one real root greater than $4$. We now observe $f(6) = -2$ and $f(7)>0$. Hence the one real root is in the interval $(6,7)$ which contains no integers.
A: $a^3−6a^2−2$ does hit the $x$-axis at $a=6.0546$, so, after placing the value of $a$ in the equation, you get the value of $a^3−6a^2−2$ to be $0.001536$ which is not equal to $0$.
