Trigonometry(XI)

taking LCM on LHS of given expression we get ,
bsin⁴x + acos⁴x = aba+b
now → bsin⁴x + a(1 - sin²x)² = aba+b
→bsin⁴x + asin⁴x - 2asin²x + a = aba+b
→(a+b)²sin⁴x - 2a(a+b)sin²x + a² =0
it's a quadratic in sin²x so solving it we get sin²x = aa+b so cos²x = ba+b

now substitue the values in sin⁸xa³ + cos⁸xb³
u will get RHS ans 1(a+b)³
it's done[1]

a more elegant solution i cn think is to writ it as sin² x = a/b cos²x -------->sin²x=a/(a+b) and cos²x=b/(a+b)
and the problem is finished[1]

mind mein calculate kar ke dekh na [1]....it is reducible to a perfect square[3][3]

1=(sin^2\alpha+cos^2\alpha )^2

From which we have asin^2\alpha=bcos^2\alpha, and the result follows.

You can use Cauchy Schwarz as has been done in

http://www.goiit.com/posts/list/trignometry-sin-4-x-sin-2-y-cos-4-x-cos-2-y-1-then-sin-4-y-sin-78900.htm

(This exact problem too was solved in this way, but unable to find that post)