Convert $$$16 r = \cos{\left(3 \theta \right)}$$$ to rectangular coordinates

Convert $$$16 r = \cos{\left(3 \theta \right)}$$$ to rectangular coordinates.

Apply the formula $$$\cos{\left(3 \alpha \right)} = \cos^{3}{\left(\alpha \right)} - 3 \sin^{2}{\left(\alpha \right)} \cos{\left(\alpha \right)}$$$ with $$$\alpha = \theta$$$: $$$16 r = - 3 \sin^{2}{\left(\theta \right)} \cos{\left(\theta \right)} + \cos^{3}{\left(\theta \right)}$$$.

From $$$x = r \cos{\left(\theta \right)}$$$ and $$$y = r \sin{\left(\theta \right)}$$$, we have that $$$\cos{\left(\theta \right)} = \frac{x}{r}$$$, $$$\sin{\left(\theta \right)} = \frac{y}{r}$$$, $$$\tan{\left(\theta \right)} = \frac{y}{x}$$$, and $$$\cot{\left(\theta \right)} = \frac{x}{y}$$$.

The input becomes $$$16 r = \frac{x^{3}}{r^{3}} - \frac{3 x y^{2}}{r^{3}}$$$.

Simplify: the input now takes the form $$$16 r^{4} - x^{3} + 3 x y^{2} = 0$$$.

In rectangular coordinates, $$$r = \sqrt{x^{2} + y^{2}}$$$ and $$$\theta = \operatorname{atan}{\left(\frac{y}{x} \right)}$$$.

Thus, the input can be rewritten as $$$- x^{3} + 3 x y^{2} + 16 \left(x^{2} + y^{2}\right)^{2} = 0$$$.

$$$16 r = \cos{\left(3 \theta \right)}$$$A in rectangular coordinates is $$$- x^{3} + 3 x y^{2} + 16 \left(x^{2} + y^{2}\right)^{2} = 0$$$A.