$\displaystyle \frac{sin^{4} \alpha}{a}+ \frac{cos ^{4}\alpha }{b}= \frac{1}{a+b}$

Prove that:

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

Thanks in advance

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- May 5th 2011, 02:00 AMandtomtrig proof
$\displaystyle \frac{sin^{4} \alpha}{a}+ \frac{cos ^{4}\alpha }{b}= \frac{1}{a+b}$

Prove that:

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

Thanks in advance - May 5th 2011, 02:38 AMabhishekkgp
- May 5th 2011, 03:35 AMandtom
Nope, I still cannot prove it :/

- May 5th 2011, 03:45 AMFernandoRevilla
- May 5th 2011, 04:00 AMandtom
so there's only 1 root

since $\displaystyle \Delta = 0$,

$\displaystyle x=sin^{2}\alpha=\frac{2a}{2(a+b)}=\frac{a}{a+b}$ - May 5th 2011, 04:46 AMabhishekkgp
- May 5th 2011, 06:44 AMandtom
OK, i got it :)

Thank you guys