system of equations.

2 ( xy + yz + zx ) = ( x + y + z ) ² - x ² - y ² - z ² = 4 - 16 = -12

Or , xy + yz + zx = -12 .

Let us construct an equation which has 3 roots , " x " , " y " , " z " , given -

x + y + z = 2 .

xy + yz + zx = -12 .

xyz = 1 .

That can be found to be -

x ³ - 2 x ² + 12 x - 1 = 0

Now , 1xy + 2 z = zxyz + 2 z ² = z1 + 2 z ²

So , we want an equation having roots " z1 + 2 z ² " , y1 + 2 y ² " , " x1 + 2 x ² " , which should be fairly easy to derive from the first equation .

So please try to solve this problem all by yourself now . If you can't , well , you know where to find me :)

xy+2z = xy + 2(2-x-y) = (x-2)(y-2)

Hence the given sum is

\sum \frac{1}{(x-2)(y-2)} = \frac{\sum x-6}{(x-2)(y-2)(z-2)}

We are given \sum x = 2 and

From Ricky's post, we have the polynomial P(t) of which x,y,z are roots, so

(x-2)(y-2)(z-2) = -(2-x)(2-y)(2-z) = - P(2)=-23

Hence the required sum should be

\frac{4}{23}

@chinmaya
the polynomial is p(t) = t³-2t²+ 12t -1

it has x,y,z as roots.

so p(t) = k(t-x)(t-y)(t-z)

since p(t) is a monic polynomial (means leading coeff of highest degree is one)

k = 1

so p(t) = (t-x)(t-y)(t-z)

that's why p(2) = (2-x)(2-y)(2-z)

thank you shubhodip!

Thanks ricky,bhatt sir ,shubhodip.