[กิตติ]

อีกวิธีแต่ยาวกว่าหน่อยหนึ่ง
$\frac{sin^4\theta }{a} +\frac{cos^4\theta}{b} = \frac{1}{a+b} $
$\frac{sin^4\theta }{a} +\frac{cos^4\theta}{b} =\frac{(1-cos^2\theta)^2}{a} +\frac{cos^4\theta}{b} $
$=\frac{1-2cos^2\theta}{a}+\frac{(a+b)}{ab}cos^4\theta $

$\frac{1-2cos^2\theta}{a}+\frac{(a+b)}{ab}cos^4\theta = \frac{1}{a+b} $
$b(a+b)(1-2cos^2\theta)+(a+b)^2cos^4\theta = ab$
$(a+b)^2cos^4\theta-2b(a+b)cos^2\theta+b^2=0$
$((a+b)cos^2\theta-b)^2=0$
$cos^2\theta = \frac{b}{a+b} $....นำไปแทน$sin^2\theta = 1-cos^2\theta$
$sin^2\theta =\frac{a}{a+b}$

$(sin^2\theta )^4 = (\frac{a}{a+b})^4 = \frac{a^4}{(a+b)^4} $
$\frac{sin^8\theta }{a^3 }= \frac{a}{(a+b)^4}$

$(cos^2\theta)^4 = (\frac{b}{a+b})^4 = \frac{b^4}{(a+b)^4}$
$\frac{cos^8\theta }{b^3 }= \frac{b}{(a+b)^4}$

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