Show that the equation of the normal at any point on the curve $x=3 \cos \theta-\cos ^{3} \theta, y=3 \sin \theta-\sin ^{3} \theta$ is $4(y \cos ^{3} \theta-x \sin ^{3} \theta)=3 \sin 4 \theta$.

(N/A) Given the parametric equations of the curve:
$x = 3 \cos \theta - \cos^3 \theta$
$y = 3 \sin \theta - \sin^3 \theta$
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = -3 \sin \theta + 3 \cos^2 \theta \sin \theta = -3 \sin \theta (1 - \cos^2 \theta) = -3 \sin^3 \theta$
$\frac{dy}{d\theta} = 3 \cos \theta - 3 \sin^2 \theta \cos \theta = 3 \cos \theta (1 - \sin^2 \theta) = 3 \cos^3 \theta$
Therefore,the slope of the tangent is:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3 \cos^3 \theta}{-3 \sin^3 \theta} = -\cot^3 \theta$
The slope of the normal is:
$m_n = -\frac{1}{dy/dx} = \frac{\sin^3 \theta}{\cos^3 \theta}$
The equation of the normal at point $(x, y)$ is:
$y - (3 \sin \theta - \sin^3 \theta) = \frac{\sin^3 \theta}{\cos^3 \theta} (x - (3 \cos \theta - \cos^3 \theta))$
Multiplying by $\cos^3 \theta$:
$y \cos^3 \theta - 3 \sin \theta \cos^3 \theta + \sin^3 \theta \cos^3 \theta = x \sin^3 \theta - 3 \cos \theta \sin^3 \theta + \sin^3 \theta \cos^3 \theta$
Simplifying:
$y \cos^3 \theta - x \sin^3 \theta = 3 \sin \theta \cos^3 \theta - 3 \cos \theta \sin^3 \theta$
$y \cos^3 \theta - x \sin^3 \theta = 3 \sin \theta \cos \theta (\cos^2 \theta - \sin^2 \theta)$
Using trigonometric identities $\sin 2\theta = 2 \sin \theta \cos \theta$ and $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$:
$y \cos^3 \theta - x \sin^3 \theta = \frac{3}{2} \sin 2\theta \cos 2\theta = \frac{3}{4} \sin 4\theta$
Thus,$4(y \cos^3 \theta - x \sin^3 \theta) = 3 \sin 4\theta$.