Question

If $\frac{{\sin }^4\alpha }{a}+\frac{{\cos }^4\alpha }{b}=\frac{1}{a+b}$ then show that $\frac{{\sin }^8\alpha }{a^3}+\frac{{\cos }^8\alpha }{b^3}=\frac{1}{(a+b{)}^3}$.

Answer 1

$\frac{{\sin }^4\alpha }{a}+\frac{{\cos }^4\alpha }{b}=\frac{1}{a+b}$
$\Rightarrow \frac{{\sin }^4\alpha }{a}+\frac{{(1-{\sin }^2\alpha )}^2}{b}=\frac{1}{a+b}$
$\Rightarrow b{\sin }^4\alpha +a(1+{\sin }^4\alpha -2{\sin }^2\alpha )=\frac{1}{a+b}$
$\Rightarrow b{\sin }^4\alpha +a+a{\sin }^4\alpha -2a{\sin }^2\alpha =\frac{1}{a+b}$
$\Rightarrow (a+b){\sin }^4\alpha -2a{\sin }^2\alpha =\frac{1}{a+b}-a$
$\Rightarrow (a+b){\sin }^4\alpha -2a{\sin }^2\alpha =\frac{1-ab-a^2}{a+b}$
$\Rightarrow {\sin }^2\alpha =\frac{a}{(a+b)},$${\cos }^2\alpha =\frac{b}{(a+b)}$

$\therefore \frac{{\sin }^8\alpha }{a^3}+\frac{{\cos }^8\alpha }{b^3}=\frac{a}{{(a+b)}^4}+\frac{b}{{(a+b)}^4}=\frac{1}{{(a+b)}^3}$