<s>[INST] 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?

 [/INST] $a^3−6a^2−2$ 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 $0$.
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