How to find a three digit number which,when reversed, becomes equal to $17$ times the square of it's cube root?

If we assume that the three digit number is of the form $100x+10y+z$,where $x \in [1,9]$ and $y,z \in [0,9]$.It seems to me that we have to solve $(x,y,z)$ from the equation $$100z+10y+x = 17 \times (100x+10y+z)^\frac{2}{3}$$

but I just can't see what to do next,any ideas?

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    $\begingroup$ There aren't that many 3-digit numbers that have cube roots :) $\endgroup$ – Srivatsan Sep 26 '11 at 2:50
  • $\begingroup$ @Srivatsan Narayanan :Indeed only five:$125,216,343,512$ and $729$ $\endgroup$ – Quixotic Sep 26 '11 at 2:58

You just need to iterate through the perfect cubes smaller than $1000$, and check which of them satisfies the condition. (The answer is $216$.)

Justification: The given condition is that $N' = 17N^{2/3}$ where $N'$ is the "reverse" of $N$. So $N^{2/3}$ is rational, and this is possible iff $N$ is a perfect cube. (More generally, if $N, a, b$ are positive integers such that $\gcd(a,b)=1$ and $N^{a/b}$ is rational, then $N$ is a perfect $b$-th power. This statement can be proved via the fundamental theorem of arithmetic.)


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