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!

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.
