We begin by expanding the Left Hand Side (LHS) of the given equation. To simplify the expression, we utilize the standard trigonometric identities for triple angles provided in the chapter:

$$\sin 3x = 3 \sin x - 4 \sin^3 x$$

and

$$\cos 3x = 4 \cos^3 x - 3 \cos x$$

First, consider the term $(\sin 3x + \sin x) \sin x$. Substituting the expansion for $\sin 3x$, we get:

$$(3 \sin x - 4 \sin^3 x + \sin x) \sin x$$

Simplifying the bracket gives:

$$(4 \sin x - 4 \sin^3 x) \sin x$$

which further expands to:

$$4 \sin^2 x - 4 \sin^4 x$$

Next, consider the term $(\cos 3x - \cos x) \cos x$. Substituting the expansion for $\cos 3x$, we get:

$$(4 \cos^3 x - 3 \cos x - \cos x) \cos x$$

Simplifying the bracket yields:

$$(4 \cos^3 x - 4 \cos x) \cos x$$

which expands to:

$$4 \cos^4 x - 4 \cos^2 x$$

Adding these two results gives the LHS as:

$$(4 \sin^2 x - 4 \sin^4 x) + (4 \cos^4 x - 4 \cos^2 x)$$

We can rearrange the terms to group like expressions:

$$4(\sin^2 x - \cos^2 x) + 4(\cos^4 x - \sin^4 x)$$

Notice that $\cos^4 x - \sin^4 x$ can be factored as:

$$(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$$

Since $\cos^2 x + \sin^2 x = 1$, this term becomes $\cos^2 x - \sin^2 x$, which is $-(\sin^2 x - \cos^2 x)$. Substituting this back, the expression becomes:

$$4(\sin^2 x - \cos^2 x) - 4(\sin^2 x - \cos^2 x)$$

which equals 0.

∴ Hence, LHS = RHS.