A literal complex system

Algebra Level 3

$\large \begin{cases}{x+y+z=2} \\ {x^2+y^2+z^2=3} \\ {xyz = 4 } \end{cases}$

If $x,y$ and $z$ are complex numbers that satsify the system of equations above, find the value of $\frac1{xy+z-1} +\frac1{yz+x-1}+\frac1{zx+y-1}$ .

-\frac13

-\frac29

2

\frac13

1 solution

A Former Brilliant Member
Dec 20, 2018

Let us just consider the denominator of the first term of the equation whose value is asked i.e. xy +z - 1.

From 1st given equation we can deduce that z -1 = 1 - x - y. hence we have xy + z - 1 = xy -x -y + 1 = (1 - x)(1 -y).

Similarly we can say that other two denominators will look like (1 - y)(1 - z) & (1 - z)(1 - x).

So our question changes from 1/(xy+z-1) + 1/(yz+x-1) + 1/(xz+y-1)

to 1/(1-x)(1-y) + 1/(1-y)(1-z) + 1/(1-z)(1-x)

now we see that a common denominator of (1-x)(1-y)(1-z) can be made , hence after multiplying and dividing each term by the needed terms we have ,

1/(1-x)(1-y) + 1/(1-y)(1-z) + 1/(1-z)(1-x) = [(1-z) + (1-x) + (1-y)]/(1-x)(1-y)(1-z) = [3 - (x+y+z)]/(1-x)(1-y)(1-z) = 1/(1-x)(1-y)(1-z)

because we have been given that x+y+z = 2 .

Now, (1-x)(1-y)(1-z) = 1 - (x+y+z) + (xy+yz+zx) - xyz = 1 - 2 + (xy+yz+zx) - 4.

Now, squaring equation 1st and then subtracting the 2nd, yields (xy+yz+zx) = 1/2

Hence, (1-x)(1-y)(1-z) = -9/2 and hence 1/(1-x)(1-y)(1-z) = -2/9