<s>[INST] Is this the shortest proof of Fermat's last theorem for exponent 3? To begin with, here we only consider the case of $n^3 + (n+1)^3 = (n+2)^3$, that is the case of consecutive integers $(x,y,z)$ in the well known equation. The general case $x^3 + y^3 = z^3$ is much more difficult to prove (and the work is yet to be done with the method outlined below).
Here we use the integer representation of a cube $n^3$ as a sequence of consecutive odd numbers. We get the sequence by writing $n^3=n\cdot n^2$. This means that we need $n$ terms with an average value of $n^2$ to represent the cube $n^3$. This representation has a big advantage over other possible representations and it is that consecutive cubes $n^3$, $(n+1)^3$, and $(n+2)^3$ have a representation that ends when the next one starts (that is they do not share any term).
For example $2^3=2\cdot 2^2= 3+5$ and $3^3=3\cdot 3^2=7+9+11$. This is a well known property of cubes when represented as a sequence of consecutive odd integers (https://en.wikipedia.org/wiki/Cube_(algebra)). This means that we can concatenate the left hand side of the equation to form a new sequence. And we need to figure out what integer this concatenated sequence represents.
It was not really difficult to do that.
So we gave an example and show why the left hand side cannot represent a cube whose value is  the right hand side of the equation.
$$6^3 + 7^3 = 8^3.$$
We write down the representation of each sequence:
$$6^3 = 6\cdot 6^2 = 31 + 33 + 35 + 37 + 39 + 41,$$ that is 6 terms with average value of $6^2=36$.
We do the same for $7^3$.
$$7^3 = 7\cdot7^2 = 43 + 45 + 47 + 49 + 51 + 53 + 55,$$ that is 7 terms with average value of $7^2=49$.
When we concatenate the two sequences into one we will call lhs, we get:
$$\text{lhs} = 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55.$$
This sequence has $(6+7)=13$ terms with average $(n^2 + (n+1)^2 +1)/2$. In our case, the lhs is therefore a sequence with 13 terms with average 43 which represents the integer $m=13\cdot 43=559=13\cdot43$. 
The right hand side is $\text{rhs}=8^3=8\cdot8^2$, so it will have 8 terms with an average of $64$. $$\text{rhs} = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71.$$ So looking at the result, we see that $\text{lhs}=13\cdot 43$ cannot be equal to $\text{rhs}=8^3$. That is  $(2n+1)\cdot (n^2 + (n+1)^2 +1)/2$ cannot be equal to $(n+2)^3$.
I am not sure if this method can be extended to the general case where $x,y$ and $z$ can take any value (not necessarily consecutive). But I think it's worth looking into it because of the simplicity of the method.
 [/INST] No.
The presumably shortest proof for the case you consider is this
Assume $$n^3+(n+1)^3=(n+2)^3. $$
Then 
$$ 0=n^3+(n+1)^3-(n+2)^3= n^3-3n^2-9n-7=(n^2-3n-9)\cdot n-7.$$
We conclude that $n\mid 7$, but neither $1^3+2^3=3^3$, nor $7^3+8^3=9^3$. $\square$
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