Factorize $ (kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) $ [closed]

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I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.

algebra-precalculus contest-math

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asked Jul 16 at 17:10

Harsh Katara

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closed as off-topic by Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel Jul 17 at 3:53

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel

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up vote
-1
down vote

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I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.

algebra-precalculus contest-math

share|cite|improve this question

asked Jul 16 at 17:10

Harsh Katara

458

closed as off-topic by Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel Jul 17 at 3:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.

algebra-precalculus contest-math

share|cite|improve this question

asked Jul 16 at 17:10

Harsh Katara

458

I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.

algebra-precalculus contest-math

share|cite|improve this question

asked Jul 16 at 17:10

Harsh Katara

458

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 17:10

Harsh Katara

458

asked Jul 16 at 17:10

Harsh Katara

458

asked Jul 16 at 17:10

Harsh Katara

458

458

closed as off-topic by Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel Jul 17 at 3:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel Jul 17 at 3:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Xander Henderson, Leucippus, Shailesh, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

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add a comment | 

2 Answers
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$$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
Name the six terms respectively as :
$$f(x,y,z,k)=A*B*C-a*b*c$$
If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.

Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.

Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.

Finally
$$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$

The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.

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answered Jul 16 at 18:47

JJacquelin

40.1k21649

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Hint
Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$

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edited Jul 16 at 17:51

answered Jul 16 at 17:27

Leox

5,0921323

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
  – user219923
  Jul 16 at 17:58
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

$$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
Name the six terms respectively as :
$$f(x,y,z,k)=A*B*C-a*b*c$$
If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.

Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.

Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.

Finally
$$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$

The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.

share|cite|improve this answer

answered Jul 16 at 18:47

JJacquelin

40.1k21649

add a comment | 

up vote
2
down vote



accepted

$$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
Name the six terms respectively as :
$$f(x,y,z,k)=A*B*C-a*b*c$$
If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.

Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.

Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.

Finally
$$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$

The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.

share|cite|improve this answer

answered Jul 16 at 18:47

JJacquelin

40.1k21649

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

$$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
Name the six terms respectively as :
$$f(x,y,z,k)=A*B*C-a*b*c$$
If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.

Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.

Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.

Finally
$$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$

The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.

share|cite|improve this answer

answered Jul 16 at 18:47

JJacquelin

40.1k21649

$$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) tag 1$$
Name the six terms respectively as :
$$f(x,y,z,k)=A*B*C-a*b*c$$
If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.

Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.

Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.

Finally
$$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)tag 2$$

The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $fneq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.

share|cite|improve this answer

answered Jul 16 at 18:47

JJacquelin

40.1k21649

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 18:47

JJacquelin

40.1k21649

answered Jul 16 at 18:47

JJacquelin

40.1k21649

answered Jul 16 at 18:47

JJacquelin

40.1k21649

40.1k21649

add a comment | 

add a comment | 

up vote
0
down vote

Hint
Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 17:27

Leox

5,0921323

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
  – user219923
  Jul 16 at 17:58
  

  
  
  
  

add a comment | 

up vote
0
down vote

Hint
Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 17:27

Leox

5,0921323

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
  – user219923
  Jul 16 at 17:58
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint
Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 17:27

Leox

5,0921323

Hint
Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$
Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,pm 1)=0.$$

share|cite|improve this answer

edited Jul 16 at 17:51

answered Jul 16 at 17:27

Leox

5,0921323

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 17:51

edited Jul 16 at 17:51

edited Jul 16 at 17:51

answered Jul 16 at 17:27

Leox

5,0921323

answered Jul 16 at 17:27

Leox

5,0921323

answered Jul 16 at 17:27

Leox

5,0921323

5,0921323

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
  – user219923
  Jul 16 at 17:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
  – user219923
  Jul 16 at 17:58
  

  
  
  
  

This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
– user219923
Jul 16 at 17:58

This is a nice hint. Do you have any advice on how one would recognize that this is the right way to approach the problem? Are there more computational ways to do it?
– user219923
Jul 16 at 17:58

add a comment |