For an equation:
$$ x-b^2/x^3+a=0 \\$$
$$ x^4-b^2+ax^3=0 \\$$
If the discriminant is positive (i.e. $> or =0$) for real roots, what is the discriminant for these equations? Can you use the discriminant to solve the inequality for $$a$$
The discriminant of
$$-b^4(256b^2 + 27a^4)$$
So the discriminant is never positive.
The discriminant is only 0 for b = 0. In this case the equation has the solutions 0 (triple root) and -a (simple root). For $b\ne0$ , the equation has 2 real solutions because the discriminant is negative. Note, that the original equation has only the solution -a in the case b=0. The multiplication with $x^3$ changes the set of solutions in this case.
With the Descartes-sign-rule we can conclude that the equation has a positive and a negative solution, if $b\ne0$.
I forgot the case a = b = 0. In this case, the only solution is 0.