Non-Funzione di Cardano

Firstly, let's convert to polar coordinates:

${r}^{3} (6\cos^{3}{\phi}) + 11 \cos^{2}{\phi} \sin{\phi} + 6 \cos{\phi} \sin^{2}{\phi} + 3 \sin^{3}{\phi} = r \cos{\phi}$

${r}^{2} (6 + 11 \tan{\phi} + 6 \tan^{2}{\phi} + \tan^{3}{\phi}) = \frac{1}{\cos^{2}{\phi}}$

Now, that was when $\theta = 0$ . When we start to rotate then:

${r}^{2} (6 + 11 \tan{(\phi - \theta)} + 6 \tan^{2}{(\phi - \theta)} + \tan^{3}{(\phi - \theta)}) = \frac{1}{\cos^{2}{(\phi - \theta)}}$

But what we really interested in are the vertical lines, lets get $x$ value by multiplying by $\cos^{2}{\phi}$ :

${x}^{2} (6 + 11 \tan{(\phi - \theta)} + 6 \tan^{2}{(\phi - \theta)} + \tan^{3}{(\phi - \theta)}) = \frac{\cos^{2}{\phi}}{\cos^{2}{(\phi - \theta)}}$

For comfort, let $\bar{\phi} = \phi - \theta$ :

${x}^{2} (6 + 11 \tan{\bar{\phi}} + 6 \tan^{2}{\bar{\phi}} + \tan^{3}{\bar{\phi}}) = \frac{\cos^{2}{(\bar{\phi} + \theta)}}{\cos^{2}{\bar{\phi}}}$

A few trigonometrical operations on the right part and:

${x}^{2} (6 + 11 \tan{\bar{\phi}} + 6 \tan^{2}{\bar{\phi}} + \tan^{3}{\bar{\phi}}) = {(\cos{\theta} - \sin{\theta} \tan{\bar{\phi}})}^{2}$

Again, purely for comfort, let $t = \tan{\bar{\phi}}$ :

${x}^{2} (6 + 11 t + 6 {t}^{2} + {t}^{3}) = {(\cos{\theta} - \sin{\theta} {t})}^{2}$

${x}^{2} = \frac{{(\cos{\theta} - \sin{\theta} {t})}^{2}}{(6 + 11 t + 6 {t}^{2} + {t}^{3})}$

${x}^{2} = \frac{{(\cos{\theta} - \sin{\theta} {t})}^{2}}{(t+1)(t+2)(t+3)}$

Now let's review what we've got: a relation between $x$ and $t$ . If we have two different values of $t$ that result in the same $|x|$ that would mean that we at least two points in the vertical line. So, we need to find such $\theta$ so that for any $x$ there'll be several values of $t$ .

Let's consider curve $\frac{1}{(t+1)(t+2)(t+3)}$ . Areas $(-\infty; -3]$ and $[-2; -1]$ are unavailable since the function is negative and ${x}^{2}$ can't be negative. What's left is intervals $(-3; -2)$ and $(-1; +\infty)$ .

The act of multiplication by ${(\cos{\theta} - \sin{\theta} {t})}^{2}$ "pulls" the curve to the x-axis line (since the function is zero when $t = \frac{1}{\tan{\theta}}$ ). So if we "pull" the "parabola" above the interval $(-3; -2)$ we'll achieve what we need. So, the condition is:

$-3 < \frac{1}{\tan{\theta}} < -2$

$\frac{1}{3} < \tan{\theta} < \frac{1}{2}$

$\arctan{\frac{1}{3}} < \theta < \arctan{\frac{1}{2}}$

Thus, $|a - b| = \arctan{\frac{1}{2}} - \arctan{\frac{1}{3}} = \arctan{\frac{1}{7}}$ , or alternatively $|a - b| = \boxed{2 \arctan{(5 \sqrt{2} - 7)}}$ .

P.S.: the unclear part is whether you can "pull" in $(-1; +\infty)$ . Seems like a no, the resulting "bump" won't be high enough to reach the minimum in $(-3; -2)$ but I got the answer and am not motivated to prove it.

P.S.S.: the last bit w/ arctan is sly, not gonna lie) . Also sorry for absence of graphs