A number puzzle asks for every real number x whose cube plus twice itself is 12, that is, x³ + 2x =…

Step 1. The derivative. $f'(x) = 3x^2 + 2 \ge 2 > 0$ for every real $x$.

Step 2. Monotonic. Since $f' > 0$ everywhere, $f$ is strictly increasing on $\mathbb{R}$. It never takes the same value twice.

Step 3. The number of roots. The equation $f(x) = 0$ has at most one real root.

Step 4. Finding it. $f(2) = 8 + 4 - 12 = 0$, so $x = 2$ is a root. Because $f$ is strictly increasing, it is the only one.

The friend. $f(1) = 1 + 2 - 12 = -9 < 0$ and $f(3) = 27 + 6 - 12 = 21 > 0$. A continuous function that changes sign must be $0$ somewhere in between, so the root lies between $1$ and $3$, and it is the only one.

Step 5. The contrast. For $g(x) = x^3 - 3x$, $g'(x) = 3(x^2 - 1)$ changes sign at $x = \pm 1$, so $g$ is not monotonic. It rises, falls and rises again, and it takes the value $0$ at $x = -\sqrt{3}$, $0$ and $\sqrt{3}$.