System of Complex Equations

First, let's do some research in $1-x-yz$ . Since $x+y+z=2$ ,so $1-x=2-x-1=y+z-1$ . then $1-x-yz=y+z-1-yz=-(y-1)(z-1)$ ,do the same to $1-y-zx$ and $1-z-xy$ . Thus, the expression, is $-(\frac{1}{(x-1)(y-1)}+\frac{1}{(x-1)(z-1)}+\frac{1}{(y-1)(z-1)}) = - \frac { (z-1)+(y-1)+(x-1)}{(x-1)(y-1)(z-1)}$ .

The numerator is $(z-1)+(y-1)+(x-1)=x+y+z-3=-1$ . As for the denominator, expanding $(x-1)(y-1)(z-1) = xyz-xy-xz-yz+x+y+z-1 =4-3+2-1=2$ . Hence, the expression equals $\frac {1}{2}$ , so the answer is $1+2=3$ .

It has been given:

$x + y + z = 2.$

Rearranging,

$y + z - 1 = 1 - x ; z + x - 1 = 1 - y ; x + y - 1 = 1 - z ;$

Therefore, denominator of given fraction changes as:

$1 - x - yz = y + z -1 - yz = - ( y - 1 ) \cdot ( z - 1 ) ;$

$1 - y - zx = z + x -1 - zx = - ( z - 1 ) \cdot ( x - 1 ) ;$

$1 - z - xy = x + y -1 - xy = - ( x - 1 ) \cdot ( y - 1 ) ;$

Hence, the given fraction becomes

$\begin{aligned} & \frac {- 1}{( y - 1 ) \cdot ( z - 1 )} + \frac {- 1}{( z - 1 ) \cdot ( x - 1 )} + \frac {- 1}{( x - 1 ) \cdot ( y - 1 )} \\ = & \frac {- ( x - 1 ) - ( y - 1 ) - ( z - 1 )}{( x - 1 ) \cdot ( y - 1 ) \cdot ( z - 1 )} \\ = & \frac {- ( x + y + z - 3 )}{ xyz + y + z + x - xy - yz - zx -1} \\ = & \frac {1}{4 + 2 - 3 - 1} = \frac {1}{2} \end{aligned}$

which is of the form $\frac {a}{b}$ . So, a = 1 and b = 2 $\Rightarrow, a + b = 3.$

[Latex Edits]

First of all let us establish some equations that are going to be useful to us
Number 1: $1-x=y+z-1 ; 1-y=x+z-1 ; 1-z=x+y-1$ Number 2: $(x-1)*(y-1)*(z-1)= xyz-xy-yz-xz+x+y+z-1 = 2$

Note that by using the equation Number 1 , the expression $\frac {1}{1-x-yz} + \frac {1}{1-y-xz} + \frac{1}{1-z-xy}$ becomes $\\ \frac {1}{(y-1)*(1-z)} + \frac {1}{(z-1)*(1-x)} + \frac{1}{(x-1)*(1-y)}$

$\\$ Now let us simplify the first two terms. $\\$ $\normalsize \frac {1}{1-x-yz} + \frac {1}{1-y-xz} = \frac {1}{(y-1)*(1-z)} + \frac {1}{(z-1)*(1-x)}$ $\\ = \large \frac{(z-1)*(2-x-y)}{(x-1)*(y-q)*(z-1)*(z-1)} = \frac{(z-1)*(z)}{2*(z-1)} = \frac{z}{2}$

Now all we have to find out is $\frac{z}{2} + \frac{1}{(x-1)(1-y)}$

Using equation Number 2 $(x-1)(1-y) = \frac {-2}{(z-1)}$

Therefore $\normalsize \frac{z}{2}+\frac{1}{(x-1)*(1-y)} = \frac{z}{2} + \frac{1}{\frac{-2}{z-1}} \\ \normalsize = \frac{z}{2} - \frac{(z-1)}{2} = \frac {1}{2} = \frac {a}{b}$

Therefore $\large a+b=\boxed 3$

After simplifying the expression, we get

$\frac{1}{1-x-yz} + \frac{1}{1-y-zx} + \frac{1}{1-z-xy} =$ $\frac{-(x^2 y z+x^2 y+x^2 z+x y^2 z+x y^2+x y z^2-x y+x z^2-x z-2 x+y^2 z+y z^2-y z-2 y-2 z+3)}{((x y+z-1) (x z+y-1) (x+y z-1))}$

Then we just use the given conditions, and we get $\frac{1}{2}$ as the value of the expression. Hence, we get $a + b = 1 + 2 = 3$

Now as $x+y+z=2$ implies $1-x=y+z-1$

So $1-x-yz=-1+y+z-zy=-(y-1)(z-1)$

So similarly,

$1-y-zx=-(x-1)(z-1)$ and $1-z-xy=-(x-1)(y-1)$

Consider the polynomial $f(a)=a^3-2a^2+3a-4$ where it have roots $x,y,z$ by Vieta's Theorem.

Now consider the polynomial;

$f(a)=((a+1)^3-2(a+1)^2+3(a+1)-4$ $=a^3+a^2+2a-2$

At which will have roots $x-1,y-1,z-1$

So we know from here by Vieta's Theorem that;

$(x-1)(y-1)(z-1)=2$ implying

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

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

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

Hence;

$\frac{1}{1-x-yz}+\frac{1}{1-y-zx}+\frac{1}{1-z-xy}=$

$(-1)(\frac{1}{(y-1)(z-1)}+\frac{1}{(x-1)(z-1)}+\frac{1}{(x-1)(y-1)})$

$=(-1)(\frac{1}{2})(x-1+y-1+z-1)$

$=(-1)(\frac{1}{2})(x+y+z-3)$

$=(-1)(\frac{1}{2})(2-3)=\frac{1}{2}$

Therefore our require value is $1+2=3$

First we notice that $x, y, z$ are roots of the cubic equation $A^3-2A^2+3A-4=0$ by Vieta's Theorem. Now we try to manipulate the algebraic expression on the left hand side so that each denominator contains only expressions of one variable, so it would be easier for us to apply the equation we have just obtained. Notice that since $yz=3-xy-zx=3-x(y+z)=3-x(2-x)=x^2-2x+3$ we can cancel out $y, z$ $\frac{1}{1-x-yz}=\frac{1}{1-x-(x^2-2x+3)}=\frac{1}{-x^2+x-2}$ . Then since $x^3-2x^2+3x-4=0$ we long divide it by the denominator, $0=x^3-3x^2+3x-4=(-x^2+x-2)(-x+1)-2$ so $-x^2+x-2=\frac{2}{-x+1}$ Now $\frac{1}{1-x-yz}=\frac{-x+1}{2}$ (Here we assumed $x\neq 1$ . In fact, if $x=1$ , we would have $y+z=1$ , $yz=4$ , $y+z+yz=3$ , adding up the first two equations and comparing with the third would give us a contradiction.)

So the left hand side would be equal to $\frac{1-x}{2}+\frac{1-y}{2}+\frac{1-z}{2}=\frac32-\frac{x+y+z}{2}=\frac12$

Thus the answer is $1+2=\boxed{3}$ .

x+y+z=2,xy+yz+zx=3,xyz=4, (x+y+z)^2=X2+y2+z2+2(xy+yz+zx), x2+y2+z2=-2, x^3 + y^3 + z^3 - 3xyz = (x+y+z) (x^2 + y^2 + z^2 - xy - yz - xz), x^3 + y^3 + z^3=2, (x + y + z) 3 = x3 + y3+ z3+ 3z2x + 3x2z + 3yz2 + 3zy2 + 3x2y + 3xy2 + 6xyz, z2x + x2z + yz2 + zy2 + x2y + xy2=-6, 1/(1−x−yz)+1/(1−y−zx)+1/(1−z−xy)=a/b, After taking lcm , (1−y−zx)(1−z−xy)+(1−x−yz)(1−z−xy)+(1−x−yz)(1−y−zx)/(1−x−yz)(1−y−zx)(1−z−xy) =a/b, simplifying and substituting the above find values we get, -2/-4=a/b=1/2, a=1,b=2, a+b=3

Since x+y+z=2, then -x=y+z-2. Then, 1-x-yz=1+y+z-2-yz=(y-1)+z(1-y)= (y-1)(1-z) . Applying the same substitution to -y and -z yields 1-y-xz= (x-1)(1-z) and 1-z-xy= (x-1)(1-y) .

Hence, \frac{1}{1-x-yz} + \frac{1}{1-y-xz} + \frac{1}{1-z-xy} ={frac{1}{(y-1)(1-z)} + \frac{1}{(x-1)(1-z)} + \frac{1}{(x-1)(1-y)}=\frac{a}{b}.

(\frac{1}{1-z}) (\frac{1}{y-1} + \frac{1}{x-1})+ \frac{1}{(x-1)(1-y)}=\frac{a}{b} (\frac{1}{1-z}) (\frac{x+y-2}{(x-1)(y-1)})+ \frac{1}{(x-1)(1-y)}=\frac{a}{b} (\frac{1}{1-z}) (\frac{-z}{(x-1)(y-1)})+ \frac{1}{(x-1)(1-y)}=\frac{a}{b} (\frac{1}{(x-1)(1-y)})((\frac{-z}{1-z}) -1)= \frac{a}{b} \frac{1}{(x-1)(1-y)(1-z)}=\frac{a}{b}

But (x-1)(1-y)(1-z) = xyz-yz-xz-xy+x+y+z-1= (4)-(3)+(2)-1=2 , then \frac{1}{(x-1)(1-y)(1-z)}=\frac{a}{b}=\frac{1}{2}.

Therefore, a+b=1+2= 3

From the equation x+y+z=3, 1-x=y+z-1.

Now 1-x-yz =y+z-1-yz =y(1-z)-1(1-z) =(y-1)(1-z)

Hence 1/(1-x-yz) = 1/((y-1)(1-z))

Similarly 1/(1-y-zx) = 1/((z-1)(1-x)) and 1/(1-z-xy) = 1/((x-1)(1-y))

(1/((y-1)(1-z)))+ (1/((z-1)(1-x))) = 1/(z-1)(1/(1-x)-1/(y-1)) =(x+y-2)/((z-1)(1-x)(y-1))

(x+y-2)/((z-1)(1-x)(y-1))+1/((x-1)(1-y)) = 1/(1-x)(y-1) ((x+y-2)/(z-1)+1) = (x+y+z-3)/((1-x)(y-1)(z-1)) = (2-3)/((1-x)(y-1)(z-1)) = 1/((x-1)(y-1)(z-1)) = 1/((xy+1-x-y)(z-1)) = 1/(xyz-(xy+xz+yz)+(x+y+z)-1) = 1/(4-3+2-1) = 1/2

plus all fraction to get a fraction and use fundamental theorem of symmetric polynomial to put it into rational polynomial of x+y+z, xy+yz+zx and xyz.

Observe that since $x=2-y-z$ , so $1-x-yz = 1 - (2-y-z) - yz = -1 +y + z - yz = -(y-1)(z-1)$ . We have similar expressions for the other two denominators, so this motivates using the substitution $a = x-1, b=y-1, c=z-1$ , and we wish to calculate $- \frac {1}{bc} - \frac {1}{ca} - \frac {1}{ab} = - \frac {a+b+c}{abc}$ .

Since $x, y, z$ are roots to the cubic polynomial $M^3 - 2M^2 + 3M - 4$ , we use the substitution $N+1=M$ to get that $a, b, c$ are roots to the cubic polynomial $(N+1)^3-2(N+1)^2+3(N+1)-4 = N^3 +N^2 +2N -2$ . Hence, by Vieta's formula, $-\frac {a+b+c}{abc} = -\frac { -\frac {1}{1} } {-\frac {-2}{1} } = \frac {1}{2}$ . Hence, $a+b =3$ .

using three given conditions , we obtain x^2 + y^2 + z^2 = -2 this implies that there must be two conjugate pairs in x ,y and z ley y and z be conjugate pairs from the new condition , we get x^3 - 2x^2 + 3x - 4 =0 solving it we get x=1.65063

putiing x in given three equations you will get y=0.174 - 1.54i z=0.174 + 1.54i

hence , [1/( 1−x−yz) +1 /(1−y−zx) +1/( 1−z−xy ) ] = 1/2

thus , 1 + 2 = 3 is the required answer.

The solution is impossible to understand. It seems like it was copied from somewhere (possibly, student's own work) without double-checking that it copied correctly. Parts seem to be simply missing.