2010/007) prove 64 {Cos^8(x) + Sin^8(x)} = cos8 x + 28cos 4x + 35

We know

(a+b)^8 = a^8 + 8 a^7b + 28 a^6b^2 + 56 a^5b^3 + 70a^4b^4 + 56 a^3b^5 + 28 a^2b^6+8ab^7+b^8

And (a-b)^8 = = a^8 - 8 a^7b + 28 a^6b^2 - 56 a^5b^3 + 70a^4b^4 - 56 a^3b^5 + 28 a^2b^6 -8ab^7+b^8

So (a+b)^8 + (a-b)^8 = 2(a^8 + b^8) + 56 (a^6b^2+a^2b^6) + 140(a^4b^4)

Putting a = e^ix and b = e^-ix

LHS = (e^ix+e^-ix)^8 + (e^ix-e^-ix)^8 = 2(e^i8x + e^-i8x)+ 56(e^4ix + e^-4ix) + 140

Or (2 cos x)^8 + (2i sin x)^8 = 2 ( 2 * cos 8x) + 56 * 2 * cos 4x + 140

Or 2^8(cos ^8x + sin ^8x) = 4 * ( cos 8x + 28 cos 4x + 35)

Or 2^6(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

Or 64(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)