An algebra problem by Medewase Timothy

The correct answer is 1988.Here is the

solution:

Multiply all, ie;

abc = (a+b)(b+c)(c+a)/(x-2)(y-2)(z-2)

abc=(a+b)(b+c)(c+a)/

(-8+(x+y+z)2-(xy+yz+zx)4+xyz)

{sub x+y+z = 2016 & xy +yz + zx = 4000}

Thus,

abc=(a+b)(b+c)(c+a)/(xyz+24)

Therefore,

xyz + 24 = (a+b)(b+c)(a+c)/(abc)

On simplification, we get;

xyz=(a+c)/b + (b+c)/a + (a+b)/c - 22

xyz= x+y+z-28

=1988.

a= (b+c) /(x-2) = (b+c)/a = x-1 = (a+b+c)/a = 1/(x-1) = a/(a+b+c)
Similarly,
b = (c+a)/(y-2)=1/(y-1) =b/(a+b+c) and
c = (a+b)/(z-2)=1/(z-1) =c/(a+b+c).
Hence
1/(x-1) + 1/(y-1) + 1/(z-1).
multiply each term by (x-1)(y-1)(z-1).
(y-1)(z-1) + (x-1)(z-1) + (x-1)(y-1) = (x-1)(y-1)(z-1)
xy + yz + xz -2(x+y+z) + 3 = xyz -(xy+yz+xz) + (x+y+z) -1
xyz - 2(xy+yz+xz) + 3(x+y+z) - 4 = 0
xyz -2(1000) + 3(2016) -4 =0
xyz=-4044