Let $$f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$$ where $$x \in R$$ and $$k \geq 1$$. Then $$f_4(x) - f_6(x)$$ equals:

We have the family of functions $$f_k(x)=\dfrac{1}{k}\bigl(\sin^k x+\cos^k x\bigr)$$ defined for every real $$x$$ and for every integer $$k\ge 1$$. In the present question we must evaluate the expression $$f_4(x)-f_6(x)$$ and then see which one of the listed numbers it equals.

First we write $$f_4(x)$$ explicitly:

$$f_4(x)=\dfrac{1}{4}\bigl(\sin^4x+\cos^4x\bigr).$$

Similarly we write $$f_6(x):$$

$$f_6(x)=\dfrac{1}{6}\bigl(\sin^6x+\cos^6x\bigr).$$

So the required difference is

$$f_4(x)-f_6(x)=\dfrac{1}{4}\bigl(\sin^4x+\cos^4x\bigr)-\dfrac{1}{6}\bigl(\sin^6x+\cos^6x\bigr).$$

In order to simplify this quantity we must express $$\sin^4x+\cos^4x$$ and $$\sin^6x+\cos^6x$$ in a more convenient form.

First we tackle the fourth powers. We recall the algebraic identity

$$(a^2+b^2)^2=a^4+b^4+2a^2b^2.$$

Choosing $$a=\sin x$$ and $$b=\cos x$$ we have $$a^2+b^2=\sin^2x+\cos^2x=1$$, so

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

Re-arranging gives

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

Next we deal with the sixth powers. We use the standard factorisation of the sum of cubes:

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

We set $$a=\sin^2x$$ and $$b=\cos^2x$$ so that

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

Since again $$\sin^2x+\cos^2x=1$$, this simplifies immediately to

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

Now we already know $$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x$$, so substituting this into the last line yields

$$\sin^6x+\cos^6x=\bigl(1-2\sin^2x\cos^2x\bigr)-\sin^2x\cos^2x=1-3\sin^2x\cos^2x.$$

With these two results in hand we return to the main difference:

$$f_4(x)-f_6(x)=\dfrac{1}{4}\Bigl[1-2\sin^2x\cos^2x\Bigr]-\dfrac{1}{6}\Bigl[1-3\sin^2x\cos^2x\Bigr].$$

We now open the brackets term by term.

First fraction:

$$\dfrac{1}{4}\bigl[1-2\sin^2x\cos^2x\bigr]=\dfrac{1}{4}-\dfrac{2}{4}\sin^2x\cos^2x=\dfrac{1}{4}-\dfrac{1}{2}\sin^2x\cos^2x.$$

Second fraction:

$$\dfrac{1}{6}\bigl[1-3\sin^2x\cos^2x\bigr]=\dfrac{1}{6}-\dfrac{3}{6}\sin^2x\cos^2x=\dfrac{1}{6}-\dfrac{1}{2}\sin^2x\cos^2x.$$

Subtracting the second from the first we get

$$$ \bigl(f_4(x)-f_6(x)\bigr)=\left(\dfrac{1}{4}-\dfrac{1}{2}\sin^2x\cos^2x\right)-\left(\dfrac{1}{6}-\dfrac{1}{2}\sin^2x\cos^2x\right). $$$

The $$-\dfrac{1}{2}\sin^2x\cos^2x$$ terms are identical and therefore cancel out, leaving only the constants:

$$f_4(x)-f_6(x)=\dfrac{1}{4}-\dfrac{1}{6}.$$

Now $$\dfrac{1}{4}=\dfrac{3}{12}$$ and $$\dfrac{1}{6}=\dfrac{2}{12}$$, so

$$f_4(x)-f_6(x)=\dfrac{3}{12}-\dfrac{2}{12}=\dfrac{1}{12}.$$

We observe that the final result contains no $$x$$, meaning the difference is a constant equal to $$\dfrac{1}{12}$$ for every real $$x$$.

Hence, the correct answer is Option B.