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Ta có $ \frac{1}{a+b}= \frac{\sin^{4}x}{a}+\frac{\cos^{4}x}{b} \ge \frac{(\sin^2x+\cos^2x)^2}{a+b}=\frac{1}{a+b}$, do dấu bằng $\frac{\sin^{4}x}{a}+\frac{\cos^{4}x}{b} \ge \frac{(\sin^2x+\cos^2x)^2}{a+b}$ tương đương với $(b\sin^2x-a\cos^2x)\ge0$.
Ta suy ra $(b\sin^2x-a\cos^2x)=0.$ hay $\frac{\sin^2x}{a}=\frac{\cos^2x}{b}\implies\frac{\sin^2x}{a}=\frac{1-\sin^2x}{b}
\implies(a+b)(\sin^2x)=a \implies \sin^2x=\frac{a}{a+b}$.
do đó $\frac{\sin^2x}{a}=\frac{1}{a+b}$, và $\frac{\cos^2x}{b}=\frac{1}{a+b}$.
tóm lại, $ \frac{\sin^{6}x}{a^{3}}+\frac{\cos^{6}x}{b^{3}}=
(\frac{\sin^2x}{a})^3+(\frac{\cos^2x}{b})^3=\boxed{\frac{2}{(a+b)^3}}$
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