Why $4a^3+27b^2<0 \iff x^3+ax+b=0$ has exactly three solutions? According to Exc. 8 Sec. 4.3 of the book Advanced Calculus by Fitzpatrick,

For numbers $a$ and $b$, prove that the following equation has exactly three solutions if and only $4a^3+27b^2<0$: $$ x^3+ax+b=0, \ \  \ \ \  x \ \text{in} \ \mathbb R.$$

Long time struggling, I can't prove neither directions. The book is pretty rudimentary and discussions like here are beyond the scope of the book. 
Let $f(x) = x^3 + ax + b$. So $f'(x) = 3x^2 + a$. Set $3x^2 + a = 0$, so that at the points $\sqrt{-a/3}$ and $-\sqrt{-a/3}$, $f'(x) = 0$. There exists exactly two maximum/minimum due to the existence of $3$ solutions. These $2$ max/min must be at $x = \sqrt{-a/3}$ and $x = -\sqrt{-a/3}$. Note that must $a<0$. Also must $f(-\sqrt{-a/3}) > 0$ and $f(\sqrt{-a/3})<0$. Putting $\pm \sqrt{-a/3}$ into $f(x)$, I fail to conclude $4a^3 + 27b^2 < 0$. As, $$-(-a/3)^{3/2}-a(-a/3)^{1/2}+b>0 \\ (-a/3)^{3/2}+a(-a/3)^{1/2}+b<0$$ results in $a<0$ and no more thing!
Please help!
 A: A third degree-polynomial $p(x)=x^3+ax+b$ has three real roots iff it has two stationary points $x_1,x_2$ and $p(x_1)\,p(x_2)<0$. Such stationary points are located at the roots of $p'(x)=3x^2+a$, hence at $\pm\sqrt{-\frac{a}{3}}$ ($a<0$ is a necessary condition, otherwise $p(x)$ is an increasing function and has only one real root). On the other hand,
$$\begin{eqnarray*} p(x_1)\,p(x_2)&=&p\left(\sqrt{-\frac{a}{3}}\right)p\left(-\sqrt{-\frac{a}{3}}\right)\\&=&\left(b+\sqrt{-\frac{a}{3}}\left(a-\frac{a}{3}\right)\right)\left(b-\sqrt{-\frac{a}{3}}\left(a-\frac{a}{3}\right)\right)\\&=&b^2+\frac{a}{3}\left(\frac{2a}{3}\right)^2=\color{red}{b^2+\frac{4a^3}{27}}.\end{eqnarray*} $$
A: Let $\alpha$, $\beta$ and $\gamma$ be the roots, then $\alpha+\beta+\gamma=0$.
The discriminant:
\begin{align*}
  \Delta &= (\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2 \\
  &=
  -27\alpha^2 \beta^2 \gamma^2-
  4(\beta \gamma+\gamma \alpha+\alpha \beta)^3 \\
  &= -27b^2-4a^3
\end{align*}
For real coefficients $a$, $b$ but with complex roots, there're two roots being conjugate.
Say $\beta=\bar{\gamma}=u+vi$, then
\begin{align*}
  \Delta &= (\alpha-u-vi)^2(\alpha-u+vi)^2(2vi)^2 \\
  &= -4v^2[(\alpha-u)^2+v^2]^2 \\
  &< 0 \\
 4a^3+27b^2 &> 0 \\
\end{align*}

For three real roots,
  $$\Delta \ge 0 \implies 4a^3+27b^2 \le 0$$

A: For the general cubic, the theory goes like this:
Let $p(x)=x^3+ax^2+bx+c$
Then $\dfrac{dp}{dx}=3x^2+2ax+b$
We need the two solutions of $\dfrac{dp}{dx}=0$, namely $r_1, r_2$, and that $p(r_1)p(r_2)\lt0$, i.e. the two stationary points are on either side of the $x$-axis.
Solving for the stationary points, $\dfrac{-2a\pm\sqrt{4a^2-12b}}{6}=\dfrac{-a\pm\sqrt{a^2-3b}}{3}$
A few preliminary results:
$r_1r_2=\dfrac b 3\qquad;\qquad r_i^2=\dfrac{2a^2-3b\pm2a\sqrt{a^2-3b}}{9}$
Putting it all together, we require:
$(r_1(r_1^2+ar_1+b)+c)(r_2(r_2^2+ar_2+b)+c)\lt0$

If we let $a=0$ we get Jack's result.
