Let  $f_k\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)$ for k = 1, 2, 3 …. Then for all $x\in R$ , the value of  $f_4\left(x\right)-f_6\left(x\right)$ is equal to

$f_k\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x);f_4\left(x\right)=\frac{1}{4}[{\sin }^{4}x+{\cos }^{4}x]$

$=\frac{1}{4}[{({\sin }^{2}x+{\cos }^{2}x)}^2-\frac{{\left(\sin 2x\right)}^2}{2}]=\frac{1}{4}[1-\frac{{\left(\sin 2x\right)}^2}{2}]$

$f_6\left(x\right)=\frac{1}{6}[{\sin }^{6}x+{\cos }^{6}x]$

$=\frac{1}{6}[{({\sin }^{2}x+{\cos }^{2}x)}^3-\frac{3}{4}{\left(\sin 2x\right)}^2]=\frac{1}{6}[1-\frac{3}{4}{\left(\sin 2x\right)}^2]$

Now $f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}-\frac{1}{6}-\frac{{\left(\sin 2x\right)}^2}{8}+\frac{1}{8}{\left(\sin 2x\right)}^2=\frac{1}{12}$