A calculus problem by Ting Sie Kim

$n = n^3 + 1 \\ n^3 - n + 1 = 0$ Since the range of $n^3 - n + 1$ stretches over all real numbers, due to the intermediate value theorem, it must cross the x axis at some point. Therefore there is at least one real solution (although there are two complex solutions as well)

Every odd degree polynomial must have a real root.

We're looking for the intersection of $y=x^3$ and $y=x-1$ . Quickly visualizing these graphs in your head shows that there must be an intersection somewhere. (It helps if you notice that, "zooming out" on the graph, $y=x^3$ looks almost like a vertical line, while $y=x-1$ is a diagonal line.)

Let that number be $x$ , we're saying that: $x=x^3 + 1$ or $x^3 - x + 1 = 0$ \ Let $f'(x) = x^3 - x + 1$ , then $f(x) = \frac{1}{4} x^4 - \frac{1}{2}x^2 + x$ , so it's easy to see that $f$ is differentiable over the reals. Therefore, due to Rolle's theorem: $\exists c \in \mathbb{R}, f'(c) = 0$ In other words, there exists a number $c$ such that: $f'(c) = c^3 - c + 1 = 0$ or $c=c^3 + 1$ So there exists a number such that it's exactly one more than it's cube $\blacksquare$ .

We know that the fundamental theorem of algebra states that there are as many real and complex roots to a polynomial as its degree. Additionally, all complex roots come in pairs. A corollary is then that all polynomials with odd degrees have at least one real root. Therefore, we know there is at least one real root that satisfies the above condition.

hey! for any number in (-infinity,1) its cube is always less than itself ex: (-5)^3 = -125 and -125 < -5