A Cubic inequality

Let $b = N \cdot a (N \in \mathbb{R^{+}})$ so that $a, b \in \mathbb{R^{+}}.$ Substituting this value into the above equation produces:

$a^3 + (Na)^3 = a - Na \Rightarrow (1+ N^3)a^3 = (1-N)a \Rightarrow a^2 = \frac{1-N}{1+N^3}.$

If $k = a^2 + 4b^2$ , then we have $k = a^2 + 4N^2a^2 = (1+4N^2)(\frac{1-N}{1+N^3})$ . Since $k$ is the sum of the squares of two positive reals, we obviously have $k = \frac{(1+4N^2)(1-N)}{1+N^3} > 0$ . This inequality is only satisfied for $0 < N < 1$ . Upon plotting $k$ as a function of $N$ we find that it has a vertical asymptote at $N = -1$ and a horizontal asymptote at $k = -4$ and satisfies the following range intervals:

$k \in (-\infty, -4)$ over $N \in (-\infty, -1)$ ;

$k \in [1, \infty)$ over $N \in (-1, 0]$ ;

$k \in (0, 1)$ over $N \in (0, 1)$ ;

$k \in (-4, 0]$ over $N \in [1, \infty)$ .

If we require $N \in (0,1)$ , then $\boxed{k < 1}$ is the correct choice.