maths question(8)

(sin^4x/a)+(cos^4/b)=(1/(a+b))
(bsin^4x + asin^4x)/(ab) = 1/ (a + b)
(a+ b)( bsin^4x + asin^4x ) =ab

兩邊除a^2b^2
(a + b) ((sin^4x/a^2b) + (cos^4x/ab^2)) = 1/ab
(sin^4x/a^2) +(cos^4x/b^2) +sin^4x/ab +cos^4x/ab =1/ab
(sin^4x/a^2) + (cos^4x/b^2) +((sin^2x + cos^2x)^2-2sin^2xcos^2x)/ab = 1/ab
(sin^4x/a^2) + (cos^4x/b^2) -2(sin^2xcos^2x)/ab = 0
(sin^2x/a +cos^2x/b)^2 =0
sin^2x/a +cos^2x/b = 0


(sin^8x/a^3) + (cos^8x/b^3)
= (sin^2x/a + cos^2x/b) (sin^4x/a^2+cos^4x/b^2-cos^2xsin^2x/ab)
=0(sin^4x/a^2+cos^4x/b^2-cos^2xsin^2x/ab)
=0

2007-07-26 20:14:11 補充：
sor...sin^2x/a - cos^2x/b =0(sin^8x/a^3) (cos^8x/b^3)= (sin^2x/a cos^2x/b) (sin^4x/a^2 cos^4x/b^2-cos^2xsin^2x/ab)=(sin^2x/a cos^2x/b)(cos^2xsin^2x/ab)=(2sin^2x/a)(sin^4x/a^2)=(2 sin^8x)/(a^3) or (2cos^8x)/(b^3)

2007-08-02 11:41:31 補充：
(sin^4x/a) (cos^4/b)=(1/(a b))'.'sin^2x/a=cos^2/b(sin^4x/a) (cos^4/b)=(sin^2x cos^2x)(sin^2x/a)=sin^2x/a=1/(a b)ans=(2 sin^8x)/(a^3)=2/(a b)^3