Let  $F_k\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)$ , when  $x\in R$  and  $k\ge 1$ , then the value of  $F_4\left(x\right)-F_6\left(x\right)$ =

$\because F_n\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)$ 
Now, $F_4\left(x\right)=\frac{1}{4}({\sin }^{4}x+{\cos }^{4}x)=\frac{1}{4}[{({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x{\cos }^{2}x]=\frac{1}{4}[1-2{\sin }^{2}x{\cos }^{2}x]$ 
and $\begin{array}{l}{F}_6\left(x\right)=\frac{1}{6}({\sin }^{6}x+{\cos }^{6}x)=\frac{1}{6}[{\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3]\\ =\frac{1}{6}\left[({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)\right]=\frac{1}{6}\left[1(1-3{\sin }^{2}x{\cos }^{2}x)\right]\end{array}$  
$\therefore F_4\left(x\right)-F_6\left(x\right)=\frac{1}{4}(1-2{\sin }^{2}x{\cos }^{2}x)-\frac{1}{6}(1-3{\sin }^{2}x{\cos }^{2}x)$ 
$=\frac{1}{12}[3-6{\sin }^{2}x{\cos }^{2}x-2+6{\sin }^{2}x{\cos }^{2}x]=\frac{1}{12}$