If real numbers x and y satisfy X / 2 ^ 5 + 5 ^ 3 + Y / 2 ^ 5 + 6 ^ 3 = 1, X / 3 ^ 5 + 5 ^ 3 + Y / 3 ^ 5 + 6 ^ 3 = 1, find x + y

If real numbers x and y satisfy X / 2 ^ 5 + 5 ^ 3 + Y / 2 ^ 5 + 6 ^ 3 = 1, X / 3 ^ 5 + 5 ^ 3 + Y / 3 ^ 5 + 6 ^ 3 = 1, find x + y

I feel that there is something wrong with the expression of LZ. I have done the original problem
There's a clever way:
Because: X / (2 ^ 5 + 5 ^ 3) + Y / (2 ^ 5 + 6 ^ 3) = 1
x/（3^5+5^3）+y/（3^5+6^3）=1
We can know that 2 ^ 5 and 3 ^ 5 are the two roots of the equation x / (Z + 5 ^ 3) + Y / (Z + 6 ^ 3) = 1 about Z respectively
The equation can be divided into two parts: Z ^ 2 + (5 ^ 3 + 6 ^ 3-x-y) Z + 5 ^ 3 * 6 ^ 3-x * 6 ^ 3-y * 5 ^ 3 = 0
From Veda's theorem, we can get: 2 ^ 5 + 3 ^ 5 = x + Y-5 ^ 3-6 ^ 3
So x + y = 2 ^ 5 + 3 ^ 5 + 5 ^ 3 + 6 ^ 3