That's kind of the inverse of what I did. I called the expression S, substituted x,y and z from the numerators with what they are supposed to be, according to xy+xz+yz=0. Then I had to expand some squares, cancel things out and finally I got S=-S-6, so 2S=-6, and hence S=-3. :-)
Good Morning, Sir, Yes, there's a 4th one, but it is too simple to mention here Therefore, please don't laugh at it because every amateur is an ordinary people We start solving the problem from the 2nd expression : Let (x + y) ÷ z + (y + z) ÷ x + (z + x) ÷ y = A Then (x + y + z) ÷ z = (x + y) ÷ z +1 Similarly (x + y + z) ÷ x = (y + z) ÷ x + 1 ; (x + y + z) ÷ y = (z + x) ÷ y + 1 By adding them together we have (x + y + z)[1 ÷ z + 1 ÷ x + 1 ÷ y] = A + 3 But [1 ÷ z + 1 ÷ x + 1 ÷ y] = 0 ⇒ A + 3 = 0, A = -3 10-24-2024
Let P1=x+y+z P2=x^2+y^2+z^2 P3=x^3+y^3+z^3 E2=xy+yz+zx E3=xyz We have E2=0 (x+y)/z+(y+z)/x+(z+x)/y = A say xy(x+y)+yz(y+z)+zx(z+x) = Axyz x^2 y + x y^2+ y^2 z + y z^2 + z^2 x + zx^2 = Axyz (x+y+z)(x^2+y^2+z^2) - (x^3+y^3+z^3) = Axyz P1 P2 - P3 = A E3 (*) One of the Newton-Girard identities is P3 - 3E3 = P1 (P2 - E2) Hence P3 = P1 P2 - P1 E2 + 3E3 Subs. for P3 in (*) P1 P2 - (P1P2 - P1 E2 + 3E3) = AE3 P1 E2 = (A+3) E3 But E2=0 so E3 = 0 or A = -3 But x, y and z must be non-zero for a solution so E3 is not 0 and A = -3
Your post is wrong, because you are missing required grouping symbols. You wrote the terms in a different order than what they were given in. You crammed your characters togethers without putting more spaces in between for better readability. You do *not* have an "oral" approach. You *still* have a *written* approach. You are not saying it. I will write a redo with some more connecting steps: (x + y)/z + (y + z)/x + (z + x)y Add 1 to each of the terms and subtract 3 at the end to balance it: [(x + y)/z + 1] + [(y + z)/x + 1] + [(z + x)y + 1] - 3 = [(x + y)/z + z/z] + [(y + z)/x + x/x] + [(z + x)y + y/y] - 3 = (x + y + z)/z + (x + y + z)/x + (x + y + z)/y - 3 = (x + y + z)(1/x + 1/y + 1/z) - 3 = (x + y + z)(xy + yz + zx)/(xyz) - 3 xy + xz + yz = 0 is given in the problem. = (x + y + z)(0) - 3 = 0 - 3 = -3 Answer
