Given, $f(x)=\sin ^6 x+\cos ^6 x$ and $x \in R$
$\begin{aligned}
& f(x)=\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3 \\
& \Rightarrow \quad f(x)=\left(\sin ^2 x+\cos ^2 x\right) \\
& \left\{\sin ^4 x+\cos ^4 x-\sin ^2 x \cdot \cos ^2 x\right\} \\
& \Rightarrow \\
& f(x)=1\left\{\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cdot \cos ^2 x\right. \\
& \left.-\sin ^2 x \cdot \cos ^2 x\right\} \\
& \Rightarrow \quad f(x)=\left\{1-3 \sin ^2 x \cdot \cos ^2 x\right\} \\
&
\end{aligned}$


$\because \quad 0 \leq \sin ^2 x \leq 1$
$\begin{array}{lc}\Rightarrow & 0 \leq(\sin 2 x)^2 \leq 1 \\ \Rightarrow & 0 \geq \frac{-3(\sin 2 x)^2}{4} \geq-3 / 4 \\ \Rightarrow & 1 \geq 1-\frac{3}{4}(\sin 2 x)^2 \geq 1-\frac{3}{4} \\ \Rightarrow & 1 \geq f(x) \geq 1 / 4 \\ \Rightarrow & f(x)[1 / 4,1] \forall x \in R\end{array}$