Think logically not mathematically [part-22]

Let the numbers be $n-1,n,n+1$ .

$\therefore (n-1)^{3}+n^{3}=(n+1)^{2}$

$n^{3}-1-3n^2+3n+n^{3}=n^{2}+2n+1$

$2n^{3}-4n^{2}+n-2=0$

$2n^{2}(n-2)+(n-2)=0$

$(2n^{2}+1)(n-2)=0$

$n-2=0\Rightarrow n=2$

Hence there is only one ordered triplet satisfying the given equation, $(1,2,3)$ .

The exists only one possibilities that is $1^3+2^3=3^2$