Find the number such that when 728 is added to it, the resulting number becomes a perfect cube whose cube root is 2 more than the cube root of the original number.
1000
Let the original number be \(n^3\), so its cube root is \(n\).
The new number is \(n^3 + 728\), and its cube root is \(n+2\), so \((n+2)^3 = n^3 + 728\).
Expanding, \(n^3 + 6n^2 + 12n + 8 = n^3 + 728\), which gives \(6n^2 + 12n - 720 = 0\), or \(n^2 + 2n - 120 = 0\).
Factoring, \((n-10)(n+12) = 0\), so the positive value is \(n = 10\).
So the original number is \(n^3 = 10^3 = 1000\).
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