Complicated Algebraic Expression

Let $a=\frac{x-y}{z}$ , $b=\frac{y-z}{x}$ , $c=\frac{z-x}{y}$ . Note that $x+y+z=0$ , so $1+a = -\frac{2y}{z}$ and $1-a=-\frac{2x}{z}$ . Similarly $1+b=-\frac{2z}{x}$ , $1-b=-\frac{2y}{x}$ , $1+c=-\frac{2x}{y}$ and $1-c=-\frac{2z}{y}$ . It follows that $(1 + a)(1 + b)(1 + c) = 1+a+b+c+ab+ac+bc+abc= -8$ and $(1 - a)(1 - b)(1 - c) = 1-a-b-c+ab+ac+bc-abc= -8.$ From this we can see that $a + b + c + abc = 0$ and $ab+ac+bc=-9$ . Now we compute the desired expression $(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = -abc \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = -(ab + ac + bc) = 9.$

Consider 3 constants $a$ , $b$ \& $c$ such that $a+b+c=0$ .

Now,

$\frac{a-b}{c} + \frac{b-c}{a} + \frac{c-a}{b}$

$= -\frac{(c-a)+(b-c)}{c} + \frac{b-c}{a} + \frac{c-a}{b}$

$= \frac{c-a}{b}-\frac{c-a}{c} + \frac{b-c}{a}-\frac{b-c}{c}$

$= (c-a) [\frac{c-b}{bc}] + (b-c) [\frac{c-a}{ac}]$

$= (c-a)(b-c)[\frac{1}{ac}-\frac{1}{bc}]$

$= \frac{(b-a)(c-b)(a-c)}{abc}$ (*)

In the original problem,it is easily noticed that $x+y+z=0$ .

Again, $(x-y)+(y-z)+(z-x)=0$ .

So in (*), putting $a=x$ , $b=y$ , $c=z$ , we get the premultiplier, which is equal to:

$\frac{(y-x)(z-y)(x-z)}{xyz}$ .

Again, putting $a=(x-y) , b=(y-z), c=(z-x)$ \& doing some simple algebraic manipulation,we prove the compatibilty of this set in (*), as in:

Consider the 2nd term, it is:

$\frac{b-c}{a} = \frac{(y-z)-(z-x)}{x-y} = \frac{x+y+z-3z}{x-y} = -3\frac{z}{x-y}$ [as $x+y+z=0$ ], which corresponds to 3rd term in the post multiplier.

[The rest simplification is similar for other two terms]

Thus after taking common -3 from this expression we get postmultiplier, which is equal to (after same manipulation with the $(b-a), (c-b), (a-c)$ terms):

$= -3 \times -3\frac{xyz}{(y-x)(z-y)(x-z)}$ .

Hence after multiplying the two expressions, what remains is a 9.

$$\frac{(x-y)}z+\frac{(y-z)}x+\frac{(z-x)}y=\frac{(x-y)}z+\frac{(y-z)}x-\frac{x-y+y-z}y$$ $$=(x-y)\left(\frac1z-\frac1y\right)+(y-z)\left(\frac1x-\frac1y\right)$$

$$=\frac{(x-y)(y-z)}{yz}-\frac{(y-z)(x-y)}{xy}=\frac{(x-y)(y-z)(x-z)}{xyz}$$

Now, $$\frac{(x-y)(y-z)(x-z)}{xyz}\cdot \frac x{y-z}=\frac{(x-y)(x-z)}{yz}=\frac{x^2+yz-x(y+z)}{yz}$$

As $$x+y+z=0, x^2+yz-x(y+z)=x^2+yz-x(-x)=2x^2+yz$$

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

So, the required sum will be$$\frac2{xyz}(x^3+y^3+z^3)+1+1+1 $$

As $$x+y+z=0, x^3+y^3+z^3=(x+y)^3+z^3-3xy(x+y)=(-z)^3+z^3-3xy(-z)=3xyz$$

So, the required sum will be$$2\cdot3+3=9 $$

Let $w=x-y$ , $u=y-z$ , $v=z-x ...$ (1) . Then,

$(\frac {x-y}{z} + \frac {y-z}{x} + \frac{z-x}{y}) \cdot (\frac {x}{z-y} + \frac{y}{z-x} + \frac {z}{x-y})= (\frac {w}{z} + \frac {u}{x} + \frac{v}{y}) \cdot (\frac {x}{u} + \frac{y}{v} + \frac {z}{w})$

On expanding, $(\frac {w}{z} + \frac {u}{x} + \frac{v}{y}) \cdot (\frac {x}{u} + \frac{y}{v} + \frac {z}{w})$

$=1+1+1 +\frac {z}{w} \cdot ( \frac {u}{x} + \frac{v}{y} ) + \frac{y}{v} \cdot (\frac {w}{z} + \frac {u}{x} ) + \frac {x}{u} \cdot (\frac {w}{z} + \frac{v}{y})$

$=3 + \frac {z}{x-y} \cdot ( \frac {y-z}{x} + \frac{z-x}{y} ) + \frac{y}{z-x} \cdot (\frac {x-y}{z} + \frac {y-z}{x} ) + \frac {x}{y-z} \cdot (\frac {x-y}{z} + \frac{z-x}{y})$ [Substituted for $w,v$ and $u$ from ... (1) ]

$=3 + \frac {z}{x-y} \cdot ( \frac {xz - yz + y^2 - x^2 }{yx} ) + \frac{y}{z-x} \cdot (\frac {yz-xy + x^2-z^2}{xz} ) + \frac {x}{y-z} \cdot (\frac {xy-xz + z^2 - y^2}{zy})$

On factorizing, $3 + \frac {z}{x-y} \cdot ( \frac {xz - yz + y^2 - x^2 }{yx} ) + \frac{y}{z-x} \cdot (\frac {yz-xy + x^2-z^2}{xz} ) + \frac {x}{y-z} \cdot (\frac {xy-xz + z^2 - y^2}{zy})$

$=3 + \frac {z}{x-y} \cdot ( \frac {(x - y)(z - (x+y))}{yx} ) + \frac{y}{z-x} \cdot (\frac {(z-x)( y-(z+x))}{xz} ) + \frac {x}{y-z} \cdot (\frac {(y-z)(x - (z+y))}{zy})$

$=3 + z \cdot ( \frac {z - (x+y)}{yx}) + y \cdot (\frac { y-(z+x)}{xz} ) + x \cdot (\frac {x - (z+y)}{zy})$ .... (2) .

Now, its given that $x=2013, y=140542, z=-142555$ .

On simple inspection it can be observed that , $-(x+y)= -(2013 + 140542) = -(142555)=z$

$\Rightarrow z= -(x+y)$ .... (3) .

Similarly, $-(z+x)=y$ , $-(z+y) = x$ .... (4) . and $x+y+z=2013 + 140542-142555=0$ ... (5) .

Substituting, using (3) and (4) in (2) .

$3 + z \cdot ( \frac {z - (x+y)}{yx}) + y \cdot (\frac { y-(z+x)}{xz} ) + x \cdot (\frac {x - (z+y)}{zy})$

$=3 + z \cdot ( \frac {z +z}{yx}) + y \cdot (\frac { y+y}{xz} ) + x \cdot (\frac {x+x}{zy})$

$=3 + \frac {2z^2}{yx} + \frac { 2y^2}{xz} + \frac {2x^2}{zy}$ [Taking out $2$ as common]

$=3 + 2(\frac {z^2}{yx} + \frac {y^2}{xz} + \frac {x^2}{zy})$ [On factorizing the denominator]

$=3 + 2(\frac {z^3}{xyz} + \frac {y^3}{xyz} + \frac {x^3}{xyz})$ $=3 + 2(\frac {z^3+y^3+x^3}{xyz})$ ... (6)

Since, $x+y+z=0$ [from (5) ]

, $x^3+y^3+z^3=3xyz$ [Standard identity]

Substituting it in equation (6) ,

$3 + 2(\frac {z^3+y^3+x^3}{xyz})=3 + 2(\frac {3xyz}{xyz})$ $=3+2(3)= 9$ .

$\therefore (\frac {x-y}{z} + \frac {z-y}{x} + \frac{z-x}{y}) \cdot (\frac {x}{z-y} + \frac{y}{z-x} + \frac {z}{x-y}) = \boxed 9$

First notice that $x+y=-z$ and eliminate $z$ . This gives, for the left hand side of the product

$\frac{y-x}{x+y}+\frac{2y+x}{x}-\frac{y+2x}{y}$

Carry out the addition

$\frac{xy(y-x)+y(x+y)(2y+x)-y(x+y)(y+2x)}{xy(x+y)}$

Careful expansion and collecting of terms of the numerator gives

$2y^3+3y^2x-3yx^2+2y^3$

Let's call this expression $f(x,y)$ and looking to factorise we notice that $f(y,y)=0$ so by the factor theorem $(y-x)$ is a factor of $f(x,y)$ . Now we can use either polynomial long divison or comparison of coefficients to find the other factor, which is $2y^2+5yx+2x^2$ this factorises to $(2x+y)(x+2y)$ .

Similar careful algebraic manipulation of the right hand side of the product with easier factorisation means that the whole product can be written

$(\frac{(y-x)(2y+x)(y+x)}{xy(x+y)})\cdot(\frac{9xy(x+y)}{(y-x)(2y+x)(2x+y)})$

Cancelling down we are left with $9$ .

Solution 1: The numbers $x,y,z$ in the statement of the problem have the property that $x+y+z=0$ . We will show that whenever this property holds, one has $\left( \frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y} \right) \cdot \left( \frac{x}{y-z} + \frac{y}{z-x} + \frac{z}{x-y} \right) = 9,$ so that the answer to the given problem is $9$ .

We first calculate $\begin{aligned} \left( \frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y} \right) \cdot \frac{x}{y-z} &=& \frac{(x-y)x}{z(y-z)}+1+\frac{(z-x)x}{y(y-z)} \\ &=& 1 + \frac{xy(x-y)+xz(z-x)}{yz(y-z)} \\ &=& 1 + \frac{x^2y-xy^2-x^2z+xz^2}{yz(y-z)} \\ &=& 1 + \frac{x^2(y-z)-x(y^2-z^2)}{yz(y-z)} \\ &=& 1 + \frac{x^2-x(y+z)}{yz} \\ &=& 1 + \frac{2x^2}{yz}, \end{aligned}$ where the last step is the only one where we used the assumption that $y+z=-x$ .

Making a cyclic shift of the variables $x,y,z$ (i.e., $x$ is replaced with $y$ , $y$ is replaced with $z$ and $z$ is replaced with $x$ ), we obtain the identity $\left( \frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y} \right) \cdot \frac{y}{z-x} = 1+\frac{2y^2}{xz}$ (still under the assumption $x+y+z=0$ ). Making one more cyclic shift yields $\left( \frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y} \right) \cdot \frac{z}{x-y} = 1+\frac{2z^2}{xy}.$ Summing the last three identities yields $\left( \frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y} \right) \cdot \left( \frac{x}{y-z} + \frac{y}{z-x} + \frac{z}{x-y} \right) = 3+2\cdot\left(\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}\right).$ Finally, we compute $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy} = \frac{x^3+y^3+z^3}{xyz}.$ Since $z=-x-y$ , we have $x^3+y^3+z^3=x^3+y^3-(x+y)^3=-3x^2y-3xy^2=-3xy(x+y)=3xyz,$ which implies that $\frac{x^3+y^3+z^3}{xyz}=3$ , and finally, $\left( \frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y} \right) \cdot \left( \frac{x}{y-z} + \frac{y}{z-x} + \frac{z}{x-y} \right) = 3+2\cdot 3=9,$ as claimed.

Solution 2: Observe that $(x-y)yx + (y-z)yz+(z-x)zx = - (x-y)(y-z)(z-x)$ , where the factorization can be obtained by using the remainder factor theorem and observing that $R(x, -x, z ) = 0$ . Hence, the first term is equal to

$\frac{ - (x-y)(y-z)(z-x) } { xyz}.$

The second term can be simplified into

$\frac { - (x+y+z) ( x^2 + y^2 + z^2 - 2xy - 2yz - 2 xz) - 9 xyz } { (x-y)(y-z)(z-x) }$

Since we have that $x+y+z = 0$ , we can conclude that the product of these 2 terms is simply 9.

Simplifying the product , I've find 9 .