I suppose you know derivatives.
Given $y(x)=ax^3+bx^2+cx+d$, find the derivative $ y'(x)=3ax^2+2bx+c$.
To find stationary points you have to solve the equation:
$$
3ax^2+2bx+c=0
$$
that is second degree. So: if this equation has no real roots then the cubic has only one real root. If you find two real solutions $x_1,x_2$ then:
if $ y(x_1)y(x_2) <0$ the cubic has three distinct real roots,
if $ y(x_1)y(x_2) =0$ the cubic has two distinct real roots, one of them double, or three coincident real roots.
if $ y(x_1)y(x_2) >0$ the cubic has only one real root.
Let $a,b,c$ be three roots of the cubic equations. We will prove that if $(2),(3)$ and $(4)$ hold, $a,b,c$ must be all positive.
Suppose the contradiction, from $(4)$ we can suppose that $a<0,b<0$ and $c>0$.
From $(2)$, we have $c> -a-b$.
From $(3)$, we have
$$ab + c(a+b)>0 \implies ab >c(-a-b)>(-a-b)^2 = a^2 +2ab+b^2$$
or $a^2+ab+b^2 <0$ (contradiction).
So, $a,b,c$ must be all positive.
Then, $(1),(2),(3)$ and $(4)$ are necessary and sufficient conditions for having all 3 positive roots.
Best Answer
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 $|a|<2$.