If (sin^4 X)/a + (cos^4 X)/b = 1Prove that 1/(a+b)^3 = (sin^8 X)/a^3 + (cos^8 Y)/b^3

Since `(sin^4 X)/a + (cos^4 Y)/b = 1`  and `1 =
((a+b)^3*(sin^8 X))/(a^3) + ((a+b)^3*(cos^8 Y))/(b^3), ` we'll equate them and we'll
check if the expression represents an identity.

`(sin^4
X)/a + (cos^4 Y)/b = ((a+b)^3*(sin^8 X))/(a^3) + ((a+b)^3*(cos^8
Y))/(b^3)`

We'll multiply the 1st fraction from the left
side by `a^2*b^3`  and the second fraction by `a^3*b^2`
.

`[a^2*b^3*(sin^4 X) + a^3*b^2*(cos^4 Y)]/(a^3*b^3) =
[b^3*(a+b)^3*(sin^8 X) + a^3*(a+b)^3*(cos^8
Y)]/(a^3*b^3)`

Since the denominators are equal, we'll have
to prove that the numerators are the same:

`[a^2*b^3*(sin^4
X) + a^3*b^2*(cos^4 Y)] = [b^3*(a+b)^3*(sin^8 X) + a^3*(a+b)^3*(cos^8
Y)]`

`` `[a^2*b^3*(sin^4 X) + a^3*b^2*(cos^4 Y)]/(a+b)^3=
b^3*(sin^8 X) + a^3*(cos^8 Y)` `{a^2*b^2*[b*(sin^4 X) + a*(cos^4 Y)]}/(a+b)^3=
b^3*(sin^8 X) + a^3*(cos^8 Y)`

Since we
cannot use the Pythagorean identity `sin^2 a + cos^2 a = 1` , because the angles x and y
are not the same, the expression above does not represent an
identity.