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I am solving the equation

$$(x^z-1)(x+y)^z (x+y-1) -x^z(x-1)((x+y)^z-1) = 0$$

(Consider $x$, $y$ and $z$ are all non zeroes)

But there is no way I could find $(x,y,z)$'s such that equality holds...

Need help. Thanks

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2  
You have unbalanced parenthesis that will be very difficult to guess the correct intent. Have you distributed all the multiplication to see what might cancel? –  abiessu Sep 30 '13 at 14:26
    
Sorry, I missed the last closing parenthesis –  Keneth Adrian Sep 30 '13 at 14:28
    
No worries, it makes sense now. What have you tried with this expression? –  abiessu Sep 30 '13 at 14:29
    
I solved the equation in terms of $x$. but not quite possible. –  Keneth Adrian Sep 30 '13 at 14:33
    
You should't say "note that..", instead you should say "consider that..". Because there is solutions for some zero variables. When you say "note that...", this mean you can conclude that, what is not the case. –  Integral Sep 30 '13 at 14:39

2 Answers 2

If $x=0$ you have $-1\cdot y^z(y-1)=0\implies y^z(1-y)=0$. Doing $y=0$ and $z\neq0$ you have infinities solutions. In short, the set $\{ (0,0,z)\in\mathbb{R}^3: \ z\neq 0 \}$ gives you solutions to this equation.

You could use similar reasoning with $y=1$ in $y^z(1-y)=0$ to get another infinite set of solutions.

Another set with nonzero solutions is $\{ (1,y,z)\in\mathbb{R}^3: \ y,z\in\mathbb{R} \}$.

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what if $x, y,$ and $z$ are nonzero. –  Keneth Adrian Sep 30 '13 at 14:37
    
$z$ is already nonzero. But of course still there is more solutions. I just showed you some( yet infinite ) of them. –  Integral Sep 30 '13 at 14:38
    
thanks @integral, what about all $x, y$ and $z$ are nonzeroes? is there one? –  Keneth Adrian Sep 30 '13 at 14:40
    
I'll think about. –  Integral Sep 30 '13 at 14:41
1  
And why is that? –  Integral Sep 30 '13 at 14:53

Note the term $(x+y)^z$ appears twice, here is an expansion of rephrasing the equation around it:

$$(x^z-1)(x+y)^z (x+y-1) -x^z(x-1)((x+y)^z-1) = 0$$

$$=(x^z-1)(x+y-1)(x+y)^z-x^z(x-1)(x+y)^z+x^z(x-1)$$

$$=(x+y)^z\left((x^z-1)(x+y-1)-x^z(x-1)\right)+x^z(x-1)$$

$$=(x+y)^z\left(y(x^z-1)-x-1+x^z(x-1)-x^z(x-1)\right)+x^z(x-1)$$

$$=(x+y)^z\left(y(x^z-1)-(x-1)\right)+x^z(x-1)$$

Noting the solution set allowed by $x=1$, now assume $x\ne 1$ and continue:

$$=(x+y)^z\left(y{x^z-1\over x-1}-1\right)+x^z$$

With a little rearrangement (and assuming $x\ne -y$),

$$\left({x\over x+y}\right)^z=1-y{x^z-1\over x-1}$$

$$\implies \left({x\over x+y}\right)^z+y{x^z-1\over x-1}-1=0$$

$$\implies {(x+y)^z-x^z\over (x+y)^z}=y{x^z-1\over x-1}$$

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If $x =1 $ , then $y^z = 0$; so $x \neq 1$ since we consider all nonzeroes $ x,y$ and $z$. –  Keneth Adrian Sep 30 '13 at 14:48
    
@KenethAdrian: I don't think that is necessarily the case, I made one minor mistake which is now corrected; $x=1$ at least has the solution $y=-1$, among others. –  abiessu Sep 30 '13 at 14:52
    
@KenethAdrian: I have corrected my algebra again, that $2$ wasn't supposed to be there. –  abiessu Sep 30 '13 at 15:29

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