Real and distinct roots of a cubic equation The real values of $a$ for which the equation $x^3-3x+a=0$ has three real and distinct roots is 
 A: The derivative cancels at $x=1$ and $x=-1$. To these correspond a maximum value of $a+2$ and a minimum value of $a-2$. In order to have three real roots, you need three $x$ intercepts; this means that you must have $a+2>0$ and $a-2<0$. So, the condition is $|a|<2$.
A: The discriminant of the polynomial is
$$108-27a^2$$
If the discriminant is positive, we have three distint real roots.
So, a must satisfy the condition |a|<2.
A: If you want them to be real and distinct...
Natural choice would be to consider :$$(x-l)(x-m)(x-n)=(x^2-(l+m)x+lm)(x-n)=x^3-(l+m+n)x^2+(lm+n(l+m))x-lmn$$
Equating the coefficients we would have : $$l+m+n=0$$
$$lm+n(l+m)=-3\Rightarrow lm-n^2=-3\Rightarrow -lmn =n(3-n^2)$$
We need to find bound for $-lmn=n(3-n^2)$... 
Can you do it by using derivative?
A: Here is a graphical approach. Note that '+ a' shifts the graph up or down.
1) Disregard '+ a'.
2) Graph $ f(x) = x^3 - 3x $
http://www.wolframalpha.com/input/?i=plot%28x%5E3+-+3x%29#
3) from the graph we see that we must find the y values at the local extrema. You did that already getting y = 2 and -2
4) It should be obvious now by inspection that a must be between -2 and 2
Final answer ,
-2 < a < 2
