Question

x^3+y^3+3xy=1 find the value of x in terms of y

Answer 1

The equation x³ + 3xy + y³ = 1 is solved in integers. Find the possible values
of x-y. 

x³ + 3xy + y³ - 1 = 0
We must factor that! That's gonna be tough: Let's try to factor this in the form: (Ax + By + C)(Dx² + Ey² + Fxy + Gx + Hy + I) = 0 where, A,B,C,D,E,F,G,H, and I are all integers Then this must be an identity x³ + 3xy + y³ - 1 = (Ax + By + C)(Dx² + Ey² + Fxy + Gx + Hy + I) The coefficient of x³ on the left is 1. The coefficient of x³ on the right is AD So AD = 1 The coefficient of x² on the left is 0. (Since there are no terms in x²) The coefficient of x² on the right is AG + CD So AG + CD = 0 The coefficient of xy on the left is 3. The coefficient of xy on the right is AH + BG + CF So AH + BG + CF = 3 The coefficient of y³ on the left is 1. The coefficient of y³ on the right is AD So BE = 1 The coefficient of x on the left is 0. (Since there are no terms in x) The coefficient of x on the right is AI + CG So AI + CG = 0 The coefficient of y on the left is 0. (Since there are no terms in y) The coefficient of y on the right is BI + CH So BI + CH = 0 The constant term on the left is -1, The constant term on the right is CI So CI = -1 AD = 1 BE = 1 AG + CD = 0 AH + BG + CF = 3 BI + CH = 0 CI = -1