<s>[INST] Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$. 
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a<  b <  c $ then $a^3 + b^3  \neq  c^3$.

My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and  $a<  b <  c $, there exists a integer $k$ such that $a=k, b=k+1, c=k+2$. Then $a^3+b^3=k^3+(k+1)^3=k^3+(k^3+3k^2+3k+1)$ and $c^3=(k+2)^3=k^3+6k^2+12k+8=(k^3+3k^2+3k+1)+(3k^2+9k+7)$. Hence, we have $k^3+(k^3+3k^2+3k+1)\neq (k^3+3k^2+3k+1)+(3k^2+9k+7)$ which implies $a^3+b^3\neq c^3$.
Does my proof valid ? And, is use Contradiction easier? If not, can anyone give me a hit or suggestion ? 
Thanks
 [/INST] You should prove one more thing: $k^3\neq3k^2+9k+7$. Since the equation $k^3-3k^2-9k-7=0$ has no integer solutions, the conclusion follows.
</s>