Strange triangle II

First calculate $\angle B$ , which is simply $\angle B=180^{\circ}-\angle A-\angle C=60^{\circ}$

By Law of Sines on $\triangle ABC$ we get:

$\dfrac{a}{\sin 40^{\circ}}=\dfrac{b}{\sin 60^{\circ}} \\ \sin 40^{\circ}=\dfrac{\sqrt{3} a}{2b}$

Now, in the identity of triple angle $\sin 3\theta=3\sin \theta-4\sin^3 \theta$ , put $\theta=40^{\circ}$ and rearrange:

$\sin 120^{\circ}=3\sin 40^{\circ}-4\sin^3 40^{\circ} \\ \dfrac{\sqrt{3}}{2}=3\sin 40^{\circ}-4\sin^3 40^{\circ} \\ \sqrt{3}=6\sin 40^{\circ}-8\sin^3 40^{\circ} \\ \sqrt{3}=6\left(\dfrac{\sqrt{3} a}{2b}\right)-8\left(\dfrac{\sqrt{3} a}{2b}\right)^3 \\ 1=\dfrac{3a}{b}-\dfrac{3a^3}{b^3} \\ b^3=3ab^2-3a^3 \\ \boxed{3a^3+b^3=3ab^2}$

If we apply Law of Sines on sides $b$ and $c$ , we don't get any of the other options.