Proof by Contradiction: (n+2)^3 = n^3 + (n+1)^3 ?

I have been working through D. L. Johnson's "Elements of Logic via Numbers and Sets" and have come across this problem, amongst others, which has me stumped.

"Prove by contradiction that the cube of the largest of three consecutive integers cannot be equal to the sum of the cubes of the other two."

Assume, then (for contradiction) that the above is not the case, i.e.

$\displaystyle (n+2)^3 = n^3 + (n+1)^3 $

for some integer $\displaystyle n $. Then

$\displaystyle \begin{array}{rcl}

n^3 + 3n^2 \cdot (2) + 3n \cdot (2)^2 + 8 & = & n^3 + n^3 + 3n^2 + 3n + 1 \\

n^3 + 6n^2 + 12n + 8 & = & 2n^3 + 3n^2 + 3n + 1 \\

n^3 - 3n^2 - 9n - 9 & = & 0

\end{array} $.

I suppose that we wish to show that this is not the case for any integer $\displaystyle n$, to provide the necessary contradiction. I'm not sure how though I might do this. Any hints much appreciated.