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

$f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}[{\sin }^{4}x+{\cos }^{4}x]-\frac{1}{6}[{\sin }^{6}x+{\cot }^{6}x]$

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

$=\frac{1}{4}[1-2{\sin }^{2}x{\cos }^{2}x]-\frac{1}{6}[1-3{\sin }^{2}x{\cos }^{2}x]=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$