Square > Cube?

Noting that $C$ is a perfect cube, let's find a $C$ with prime factor 2, which is the smallest prime number:

• If $C=2^6=\big(2^2\big)^3=64,$ $N$ can't be a 3-digit number since $N<C.$ So it doesn't work.
• If $C=2^9=\big(2^3\big)^3=512,$ the number $N=2^7=128$ makes $C=512$ the smallest perfect cube divisible by $N,$ and $S=2^8=256$ the smallest perfect square divisible by $N.$ However, since the condition $C<S$ is not fulfilled, it doesn't work either.
• If $C=2^9=\big(2^3\big)^3=512,$ the number $N=2^8=256$ makes $C=512$ the smallest perfect cube divisible by $N,$ and $S=2^{10}=1024$ the smallest perfect square divisible by $N.$ In addition, in this case, the condition $N<C<S$ is also fulfilled. So, $(N, C, S)=(256, 512, 1024)$ is a candidate for our solution.

Now, let's find a $C$ with prime factors 2 and 3, which are the smallest two prime numbers:

• If $C=2^3 3^3=\big(2\times 3\big)^3=216,$ the number $N=2^3 3^2=72$ makes $C=216$ the smallest perfect cube divisible by $N,$ and $S=2^4 3^2=144$ the smallest perfect square divisible by $N.$ However, neither is $N$ a 3-digit number nor is the condition $C<S$ fulfilled, so it doesn't work.
• If $C=2^3 3^3=\big(2\times 3\big)^3=216,$ the number $N=2^2 3^3=108$ makes $C=216$ the smallest perfect cube divisible by $N,$ and $S=2^2 3^4=324$ the smallest perfect square divisible by $N.$ Obviously, the condition $N<C<S$ is fulfilled. So, $(N, C, S)=(108, 216, 324)$ works. In addition, it works better than $(N, C, S)=(256, 512, 1024)$ because $N=108$ here is smaller than 256.

Therefore, $S-C-N=324-216-108=0.$ $_\square$

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Note: $N=10^2=100$ might look attractive. But in this case, $C=10^3=1000$ and $S=2^2 10^2=400,$ which fails to satisfy the condition $C<S.$

Used hit and trial method. Started with 100, 125 and then 108 given $6^{3}$ was intuitively attractive.