Answer
The equation $~~x^3+e^x = 0~~$ has exactly one real root.
Work Step by Step
$x^3+e^x = 0$
Let $f(x) = x^3+e^x$
The function $f(x) = x^3+e^x$ is continuous and differentiable for all $x$.
When $x=-1$, then $(x^3+e^x) \lt 0$
When $x=1$, then $(x^3+e^x) \gt 0$
Therefore, by the Intermediate Value Theorem, there is a number $c$ in the interval $(-1,1)$ such that $f(c) = 0$.
Thus the equation has at least one real root.
Let's assume that the equation has at least two real roots $a$ and $b$. Then $f(a) = f(b) = 0$. According to Rolle's Theorem, there is a number $k$ such that $f'(k) = 0$.
We can try to find $k$:
$f'(x) = 3x^2+e^x = 0$
$x^2 = -\frac{e^x}{3}$
However, $x^2 \geq 0$ for all $x$ while $-\frac{e^x}{3} \lt 0 $ for all $x$. There is no number $x$ such that $f'(x) = 0$. Therefore, the assumption that the equation has at least two real roots is incorrect.
Therefore, the equation has exactly one real root.