Ta có ${\sin ^8}x + {\cos ^8}x = \dfrac{a}{{64}} + \dfrac{b}{{16}}\cos 4x + \dfrac{c}{{64}}\cos 8x$ với $a,b \in \mathbb{Q}$. Khi đó $a - 5b + c$ bằng

${\sin ^8}x + {\cos ^8}x$$ = {\left( {{{\sin }^4}x + {{\cos }^4}x} \right)^2} - 2{\sin ^4}x.{\cos ^4}x$$ = {\left( {1 - 2{{\sin }^2}x.{{\cos }^2}x} \right)^2} - \dfrac{1}{8}{\sin ^4}2x$

$ = {\left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right)^2} - \dfrac{1}{8}{\sin ^4}2x$$ = 1 - {\sin ^2}2x + \dfrac{1}{8}{\sin ^4}2x$$ = 1 - \dfrac{{1 - \cos 4x}}{2} + \dfrac{1}{8}{\left( {\dfrac{{1 - \cos 4x}}{2}} \right)^2}$

$ = 1 - \dfrac{{1 - \cos 4x}}{2} + \dfrac{1}{{32}}\left( {1 - 2\cos 4x + \dfrac{{1 + \cos 8x}}{2}} \right)$$ = \dfrac{{35}}{{64}} + \dfrac{7}{{16}}\cos 4x + \dfrac{1}{{64}}\cos 8x$

$ \Rightarrow a = 35$, $b = 7$, $c = 1$$ \Rightarrow a - 5b + c = 1$.