If sin4x/a + cos4x/b= 1/(a+b), then prove that sin8x/a3 + cos8x/b3=2/(a+b)3?

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

=> sin^4 x/a + (cos^2 x)^2/b = 1/(a + b)

=>( sin^2 x)^2/a + (1 - sin^2 x)^2/b = 1/(a + b)

Let, sin^2 x = k

k^2/a + (1 - k)^2/b = 1/(a + b)

=> [bk^2 + a(1 - k)^2]/ab = 1/(a+ b)

=> [bk^2 +a(1 - 2k + k^2)]/ab = 1/(a + b)

=> [bk^2 + a - 2ak + ak^2]/ab = 1/(a + b)

=> [k^2(a + b) - 2ak + a]/ab = 1/(a + b)

=> k^2(a + b)^2 - 2a(a + b)k + a(a + b) - ab = 0

=> k^2(a+ b)^2 - 2a(a + b)k + a^2 + ab - ab = 0

=> k^2(a + b)^2 - 2a(a + b)k + a^2 = 0

=> {k(a + b)}^2 - 2.k(a + b).a + (a)^2 = 0

=> [k(a+ b) - a]^2 = 0

=> k(a + b) - a = 0

=> k(a + b) = a

=> k = a/(a + b)

Hence,

sin^2 x = a/(a+ b)

cos^2 x = 1 - a/(a + b) =[a + b - a] /(a + b) = b/(a+ b)

Now,
sin^8 x = (sin^2 x)^4 = [a/(a + b)]^4 = a^4/(a + b)^4

=> sin^8 x/a^3 = a^4/(a + b)^4.a^3 = a/(a + b)^4

Similarly, cos^8 x = (cos^2 x)^4 = [b/(a + b)]^4
= b^4/(a + b)^4

=> cos^8 x/b^3 =b^4/(a + b)^4.b^3 = b/(a + b)^4

Now, sin^8 x/a^3 + cos^8 x/b^3

= a/(a+ b)^4 + b/(a + b)^4

= (a + b)/(a+ b)^4

= 1/(a + b)^3<==ANSWER

Hope this helps u:)