Question

if x²=y+z, y²=z+x, z²=x+y find {1/(x+1)}+{1/(y+1)}+{1/(z+1)}

Answer 1

x²=y+z, y²=z+x, z²=x+y The three equations are symmetrical, so anything we discover about any pair of variables, it will also be true of any other pair of variables. Solve the first two for z. x²-y=z y²-x=z x²-y = y²-x, since both equal to z x²-y² = -x+y (x-y)(x+y) = -(x-y) We will see if we can divide by x-y We will first assume x-y ≠ 0. Then we can divide both sides by x-y and get x+y = -1 But since z²=x+y, then that would mean z²=-1 and z would be ±i. Then by the symmetry x and y would also be ±i but then x²=y+z would give -1=i+i, -1=i-i, -1=-i+i or -1=-i-i, none of which are possible. Therefore we cannot divide both sides of (x-y)(x+y) = -(x-y) by x-y, and therefore the assumption x-y≠0 was false and therefore x-y=0 and x=y. By symmetry x=y=z, so all three equations amount to x²=x+x x²=2x x²-2x = 0 x(x-2) = 0 x=0, x=2 By symmetry then x=y=z=0 or x=y=z=2 If x=y=z=0 then = = = = 3 If x=y=z=2 then = = = 1 Answer: 3 or 1