For which $a, b$ in $(-1,1)$ does $\frac{1}{3}x^3-a^2x+b=0$ have three real solutions?

Question

I know $\frac{1}{3}x^3-a^2x+b=0$ will always have one solution as $\frac{-b}{x} = \frac{1}{3}x^2-a^2$ will always have one intersection in the upper two quadrants of the coordinate system.

But I can't really make out for the bottom two quadrants.

Answer 1

Hint: The number of a real solutions for the equation $x^3+3px+2q=0$ is in relation to the sign of discriminant $D=q^2+p^3$. Above equation will have 3 real distinct solutions if $D<0$!

Answer 2

I assume you mean "distinct real solutions for real $x$", since otherwise the question is boring: any degree three polynomial has three complex roots counted with multiplicity.

The discriminant of this cubic is $\frac{1}{3}(4a^6 - 9b^2)$. The cubic has three distinct real roots if the discriminant is positive, which requires $4a^6 > 9 b^2$. This region is indicated in this plot.

(Note that the boundary of the shaded regions is not suitable choices of $(a,b)$. For instance, $(a,b) = (0,0)$ does not yield three distinct solutions.)