1. Simplify the trigonometric expression

We need to solve

$$\frac{3\cos 2x+\cos^3 2x}{\cos^6 x-\sin^6 x}=x^3-x^2+6.$$

First simplify the denominator:

$$\cos^6 x-\sin^6 x=(\cos^2 x)^3-(\sin^2 x)^3.$$

Using $a^3-b^3=(a-b)(a^2+ab+b^2)$,

$$\cos^6 x-\sin^6 x=(\cos^2 x-\sin^2 x)(\cos^4 x+\cos^2 x\sin^2 x+\sin^4 x).$$

Now,

$$\cos^2 x-\sin^2 x=\cos 2x.$$

Also,

$$\cos^4 x+\sin^4 x=(\cos^2 x+\sin^2 x)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x.$$

So,

$$\cos^4 x+\cos^2 x\sin^2 x+\sin^4 x=1-\sin^2 x\cos^2 x.$$

Using

$$\sin^2 2x=4\sin^2 x\cos^2 x,$$

we get

$$1-\sin^2 x\cos^2 x=1-\frac{\sin^2 2x}{4}=1-\frac{1-\cos^2 2x}{4}=\frac{3+\cos^2 2x}{4}.$$

Hence,

$$\cos^6 x-\sin^6 x=\cos 2x\cdot \frac{3+\cos^2 2x}{4}.$$

Now simplify the numerator:

$$3\cos 2x+\cos^3 2x=\cos 2x(3+\cos^2 2x).$$

Therefore,

$$\frac{3\cos 2x+\cos^3 2x}{\cos^6 x-\sin^6 x} = \frac{\cos 2x(3+\cos^2 2x)}{\cos 2x\cdot \frac{3+\cos^2 2x}{4}}=4,$$

provided the denominator is nonzero.

So the equation becomes

$$4=x^3-x^2+6.$$

That is,

$$x^3-x^2+2=0.$$

2. Solve the cubic equation

We factor:

$$x^3-x^2+2=0.$$

Try rational roots $\pm 1, \pm 2$.

At $x=-1$,

$$(-1)^3-(-1)^2+2=-1-1+2=0.$$

So $(x+1)$ is a factor.

Divide:

$$x^3-x^2+2=(x+1)(x^2-2x+2).$$

Now solve

$$x^2-2x+2=0.$$

Discriminant:

$$\Delta =(-2)^2-4(1)(2)=4-8=-4<0.$$

So this quadratic has no real roots.

Thus the only real solution is

$$x=-1.$$

3. Check domain restriction

We must ensure the original denominator is not zero:

$$\cos^6 x-\sin^6 x\neq 0.$$

At $x=-1$,

$$\cos^2(-1)=\cos^2 1,\qquad \sin^2(-1)=\sin^2 1,$$

which are not equal, so denominator is nonzero. Thus $x=-1$ is valid.

4. Find the sum of all real solutions

There is only one real solution, namely $x=-1$.

Hence the sum of all real solutions is

$$-1.$$

5. Compare with stored answer

Derived answer: Option D, $-1$.

This matches the stored correct answer.