$If $x,y,z$ are real no. such that $\left\{\begin{array}{c} x+y+z=2\\ x^2+y^2+z^2=16\\ xyz=1\end{array}\right.$ .\\\\\\ Then Calculate value of $\displaystyle \frac{1}{xy+2z}+\frac{1}{yz+2x}+\frac{1}{zx+2y}=$
12 ( 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 :)
341xy+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}
21@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)
