If  $f_n\left(x\right)=\frac{1}{n}\left({\sin }^{n}x+{\cos }^{n}x\right)\text{ then }\frac{1}{f_4\left(x\right)-f_6\left(x\right)}$  must be equal to 

$Given : f_n\left(x\right)=\frac{1}{n}\left({\sin }^{n}x+{\cos }^{n}x\right)$

$$\begin{gathered} f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}\left({\sin }^{4}x+{\cos }^{4}x\right)-\frac{1}{6}\left({\sin }^{6}x+{\cos }^{6}x\right) \\ =\frac{1}{4}\left[{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^2\left(x\right){\cos }^2\left(x\right)\right]-\frac{1}{6}\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{4}x+{\cos }^{4}x-s{\mathrm{in}}^2\left(x\right){\cos }^2\left(x\right)\right) \\ =\frac{1}{4}-\frac{1}{2}{\sin }^2\left(x\right){\cos }^2\left(x\right)-\frac{1}{6}\left[1-3{\sin }^2\left(x\right){\cos }^2\left(x\right)\right] \\ =\frac{1}{4}-\frac{1}{2}{\sin }^2\left(x\right){\cos }^2\left(x\right)-\frac{1}{6}+\frac{1}{2}{\sin }^2\left(x\right){\cos }^2\left(x\right) \\ =\frac{1}{4}-\frac{1}{6} \\ =\frac{1}{12} \\ So , \frac{1}{f_4\left(x\right)-f_6\left(x\right)}=12 \end{gathered}$$