Matematik

(5x + 8cos(0)³)·(5x - 8cos(0)³) =
(x^-1 - 5x³)·(x^-1 + 5x³) =
(8b² - 5c^-2)·(3a + 8a²) =
(x + 5b)·(cos(π/2)³ - 7a^-2) =
(5b³ - 3cos(0)²)·(6a^-2 - c^-1) =
(cos(0) + 8cos(0)^-2)·(cos(0) - 8cos(0)^-2) =
(8cos(π)^-2 + 7b^-2)² =
(8x³ - 8x³)² =
(6x^-1 + 7cos(2π/3)^-2)² =
(5a² + 8cos(π)^-2)² =
(6b² + c³)·(3b^-1 + 7x^-1) =
(3cos(2π/3)³ - 6b)² =
(8cos(π)² - 7b)² =
(3b + 8a³)·(8cos(2π/3)³ + 8a^-1) =
(3b - 3b³)² =
(8cos(π)³ + 7b)² =
(a^-1 - 5a^-2)² =
(cos(2π/3)² + 5c)·(a³ - 5x^-1) =
(3c^-1 - c)·(6a^-1 + 3x²) =
(6c - 7cos(π/2)²)·(cos(π)² + 6cos(2π/3)^-2) =
(x^-2 + 7cos(2π/3)³)² =
(7x + 8x^-2)·(7x - 8x^-2) =
(6a - 3b²)·(8x³ + 7b) =
(6cos(0)² + 3a^-2)² =
(3b³ - 7cos(2π/3)^-1)·(5b² + 6c³) =