Algebraic Teaser

Firstly we can see that the equations are symmetric in x,y,z. So all x,y,z will have same set of solutions. Now We can assume a cubic equation whose roots are x,y,z now sum of roots = x+y+z=9 now since the above equations are symmetric in x,y,z so sum of all possible values of x = sum of roots of the cubic equation = 9

x+y+z = 9

(x+y+z)2 = 81 [Squaring both sides]

x2+y2+z2 + 2(xy + yz + xz) = 81

29 + 2(xy + yz + xz) = 81 [Substituting the value of x2+y2+z2 from equation (ii)]

xy+xz+yz=(81-29)/2 = 26

yz=26 - xy - zx

yz=26-x(y+z)

yz=26-x(9-x) [Substituting the value of x from equation (i)]

yz=26 – 9x +x2 …………………(iv)

Now, we expand y3 + z3 in the (iii) equation,

x3 + (y+z)(y2 + z2 – yz) = 9

x3 + (9-x)[29-x2 –(26 – 9x + x2)] {Substituting the value of yz from (iv)]

After solving this we get, 3x3 – 27x2 + 78x + 18 = 0

And we know that the sum of the zeroes of a cubic polynomial = -b/a Here a=3 and b= -27

-b/a = -(-27)/3 = 9