Every perfect cube is the difference of two perfect squares? How would you prove this without induction? I know that one easy way is using Al Kharchi's principle (namely that $1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2$), but are there other ways? Thanks! 

If $a^2-b^2=n^3=n^2\cdot n$
we can set $a+b= n^2$ and $a-b=n$
so that $a=\frac{n^2+n}2=\frac{n(n+1)}2$ which is an integer as $n(n+1)$ is even
Similarly, $b=\frac{n^2-n}2=\frac{n(n-1)}2$
More generally, if $a^2-b^2=n^{k+1}=n^k\cdot n$ for integer $k\ge2$
we can set $a+b= n^k$ and $a-b=n$  so that $a=\frac{n^k+n}2=\frac{n(n^{k-1}+1)}2$  
Observe that $n$ and $n^{k-1}+1$ have opposite parities, making the product even 
Similarly,  $b$ can be handled
