For any a, b, x, y greater than 0, prove that: 2/3 (tan^-1{(3ab^2-a^3)/(b^3-3a^2b)} + 2/3 {(tan^-1(3xy^2-x^3)/(y^3-3x^2y)} = tan^-1{(2αβ/(α^2-β^2)}, where α = -ax+by, β = bx+ay.