How do I show that there exists a real number that equals its cube plus its square plus 1? How do I show that there exists a real number that equals its cube plus its square plus 1?
I was thinking  $x = x^3+x^2+1$ then solve for $x$?
 A: To make things as simple as possible: you want $x$ such that $x = x^3 + x^2 + 1$, that is $x^3 + x^2 - x + 1 = 0$. And every cubic with real coefficients has a real root (because it has different signs at $x$ and $-x$ for large enough $x$).
A: Let $f(x) := x^3 + x^2 -x + 1$. Consider the following,
$$\mathrm{sgn}\left(\lim_{x \to \infty} \quad f(x) \right),$$
and,
$$\mathrm{sgn}\left(\lim_{x \to -\infty} \quad f(x) \right).$$
What can you conclude?
A: Another way is to show that it is continuous, and then find a value of the curve $y=x^3+x^2+1$ below the $y=x$ line and another point above the $y=x$ line. 
A: If a polynomial has real coefficients then its solutions will be either real or come in complex conjugate pairs. That implies that every polynomial of odd degree with real coefficients has at least one real root.
The reason that the non-real solutions must come in conjugate pairs is because they must have the same symmetry as the polynomial: The polynomial does not change when you switch i with -i so the solution set cannot change when you do that either.

A polynomial (with real coefficients) is a statement in the language of rings that defines a set of points $X$ in $\mathbb C$. These sets are called $\mathbb R$ definable and the symmetries of $\mathbb C$ that fix $\mathbb R$ are relevant, let $\sigma$ generate them then $x \in X \iff x \in \sigma(X)$.
