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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

given:

$xyz=\dfrac {a}{2}---(1)$

$x^2+y^2+z^2=a^2+6---(2)$

$x+y+z=a---(3)$

find:$\dfrac{1}{xy+az}+\dfrac{1}{yz+ax}+\dfrac{1}{zx+ay}=?$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

given:

$xyz=\dfrac {a}{2}---(1)$

$x^2+y^2+z^2=a^2+6---(2)$

$x+y+z=a---(3)$

find:$\dfrac{1}{xy+az}+\dfrac{1}{yz+ax}+\dfrac{1}{zx+ay}=?$

- Mar 31, 2013

- 1,349

given:

$xyz=\dfrac {a}{2}---(1)$

$x^2+y^2+z^2=a^2+6---(2)$

$x+y+z=a---(3)$

find:$\dfrac{1}{xy+az}+\dfrac{1}{yz+ax}+\dfrac{1}{zx+ay}=?$

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

Last edited:

- Thread starter
- #3

- Jan 25, 2013

- 1,225

= 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