Nếu biết $\dfrac{{{{\sin }^4}\alpha }}{a} + \dfrac{{{{\cos }^4}\alpha }}{b} = \dfrac{1}{{a + b}}$ thì biểu thức $A = \dfrac{{{{\sin }^8}\alpha }}{{{a^3}}} + \dfrac{{{{\cos }^8}\alpha }}{{{b^3}}}$ bằng

Đặt \({\cos ^2}\alpha  = t \Rightarrow \dfrac{{{{\left( {1 - t} \right)}^2}}}{a} + \dfrac{{{t^2}}}{b} = \dfrac{1}{{a + b}}\)

\( \Leftrightarrow b{\left( {1 - t} \right)^2} + a{t^2} = \dfrac{{ab}}{{a + b}}\)\( \Leftrightarrow a{t^2} + b{t^2} - 2bt + b = \dfrac{{ab}}{{a + b}}\)\( \Leftrightarrow \left( {a + b} \right){t^2} - 2bt + b = \dfrac{{ab}}{{a + b}}\)\( \Leftrightarrow {\left( {a + b} \right)^2}{t^2} - 2b\left( {a + b} \right)t + {b^2} = 0\)\( \Leftrightarrow t = \dfrac{b}{{a + b}}\)

Suy ra ${\cos ^2}\alpha  = \dfrac{b}{{a + b}};\,{\sin ^2}\alpha  = \dfrac{a}{{a + b}}$

Vậy: \(\dfrac{{{{\sin }^8}\alpha }}{{{a^3}}} + \dfrac{{{{\cos }^8}\alpha }}{{{b^3}}} = \dfrac{a}{{{{\left( {a + b} \right)}^4}}} + \dfrac{b}{{{{\left( {a + b} \right)}^4}}} = \dfrac{1}{{{{\left( {a + b} \right)}^3}}}.\)