If real numbers x and y satisfy the following conditions: x ^ (3 ^ 3 + 4 ^ 3) + y ^ (3 ^ 3 + 6 ^ 3) = 1, x ^ (5 ^ 3 + 4 ^ 3) + y ^ (5 ^ 3 + 6 ^ 3) = 1, then x + y = The answer is 432, but do not know the process, please be more detailed, please!

If real numbers x and y satisfy the following conditions: x ^ (3 ^ 3 + 4 ^ 3) + y ^ (3 ^ 3 + 6 ^ 3) = 1, x ^ (5 ^ 3 + 4 ^ 3) + y ^ (5 ^ 3 + 6 ^ 3) = 1, then x + y = The answer is 432, but do not know the process, please be more detailed, please!

Let x (3 ^ 3 + 4 ^ 3) = cos ^ 2 (θ), y (3 ^ 3 + 6 ^ 3) = sin ^ 2 (θ)
We get: x = (3 ^ 3 + 4 ^ 3) × cos ^ 2 (θ), y = (3 ^ 3 + 6 ^ 3) × sin ^ 2 (θ)
Bring in the second formula: x ^ (5 ^ 3 + 4 ^ 3) + y ^ (5 ^ 3 + 6 ^ 3) = 1
The equation about θ is obtained
(3^3+4^3)÷(5^3+4^3)×cos^2(θ）+（3^3+6^3)÷(5^3+6^3)×sin^2（θ）=1
The simultaneous equation: cos ^ 2 (θ) + sin ^ 2 (θ) = 1
Cos ^ 2 (θ) and sin ^ 2 (θ) are obtained
You can get x, y, and then x + y