Solving 3 Equations for x, y & z in R

[Albert1]

$a,x,y,z\in R,\,\, a\neq 0$
given:
$xyz=\dfrac {a}{2}---(1)$
$x^2+y^2+z^2=a^2+6---(2)$
$x+y+z=a---(3)$
find:$\dfrac{1}{xy+az}+\dfrac{1}{yz+ax}+\dfrac{1}{zx+ay}=?$

 

[kaliprasad]

$a,x,y,z\in R,\,\, a\neq 0$
given:
$xyz=\dfrac {a}{2}---(1)$
$x^2+y^2+z^2=a^2+6---(2)$
$x+y+z=a---(3)$
find:$\dfrac{1}{xy+az}+\dfrac{1}{yz+ax}+\dfrac{1}{zx+ay}=?$

Spoiler

we have
xy + az = xy + z (x + y + z) = (z+x)(z+y) = (a-y)(a-x)
similarly
yz + ax = (a-y)(a-z)
zx + ay = (a-z)(a-x)

so 1/(xy + az) + 1/(yz + ax )+ 1/(zx + ay)}
= 1/ (a-y)(a-z)+ 1/ (a-y)(a-x) + 1/ (a-z)(a-x)
= (3a – x – y – z)/)(a-y)(a-z)(a-x))

Numerator = 3a – a ( from (3)) = 2a
Denominator = $a^3- (x+y+z)a + (xy + yz + zx) a – xyz$
= $a^3- a. a^2 + (xy + yz + zx) a – xyz$
= (xy + yz + zx) a – xyz
We have $2(xy + yz+ zx) = (x+y+z)^2 – (x^2 + y^2 + z^2) = a^2 – (a^2 + 6) = - 6$
Or xy + yz + zx = - 3
So denominator = -3a – a/2 = - 7a/2
So value = - 4/7

 

[Albert1]

Denominator =a^3- (x+y+z)a + (xy + yz + zx) a – xyz
= a^3- a. a^2 + (xy + yz + zx) a – xyz
= (xy + yz + zx) a – xyz
a typo(in red)
it should be :
Denominator =a^3- (x+y+z)a^2 + (xy + yz + zx) a – xyz
=a^3- a. a^2 + (xy + yz + zx) a – xyz
= (xy + yz + zx) a – xyz

in deed ,a very nice solution :)