Is any number exactly 2 less than its cube?

This is what I did:

$\displaystyle x=x^3 - 2$. I graphed the equation and came up with x = 1.521. Is this correct?

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- Oct 24th 2009, 05:48 PMlive_laugh_luv27Cubic function
Is any number exactly 2 less than its cube?

This is what I did:

$\displaystyle x=x^3 - 2$. I graphed the equation and came up with x = 1.521. Is this correct? - Oct 24th 2009, 05:50 PMProve It
- Oct 24th 2009, 09:45 PMmr fantastic
No. The number you've found is not EXACTLY 2 less than its cube.

The question as stated requires you to establish the existence (or otherwise) of such a number, not to actually find the number. Existence is simple to prove since every cubic equation has at least one real root ....