Question

Prove that for all \(n\in \mathbb{N}\), there exists \(x\in \mathbb{R}\) for which \(x^3 = n\).
How to start?

Answer 1

@skullpatrol

Answer 2

there are several ways to approach this i think

first thought that occurs to me is calculus :
Notice that the funciton \(\large f(x) = x^3-n\) is increasing for all \(\large x\in \mathbb{R}\)

Answer 3

You may use use fundamental theorem of algebra combined with the fact that the complex zeroes occur in conjugate pairs to conclude that there exists atleast one real zero to x^3-n.

Answer 4

But calculus is in the later part of this course, I don't think I can use it...