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Is there a number that is exactly 4 more than its cube?
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x - x^3 = 4
Well, lets call such a number k. That would imply that \[k^3 + 4 = k\]\[\implies k^3 - k + 4 = 0\] So now we just need to find the roots of this polynomial.
It doesn't look like it'll be a terribly pretty number, but since it's an odd degree polynomial it must have at least one root, so the number does exist.
It won't. All three roots are irrational.
We can find the exact solutions for this equations. They involve cube roots of quadratic surds.
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