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It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1

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It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1  [#permalink]

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New post Updated on: 22 Apr 2019, 00:38
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Question Stats:

28% (02:52) correct 72% (02:25) wrong based on 29 sessions

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It is given that \(x^2+y^2=z+1\), \(y^2+z^2=x+1\) and \(z^2+x^2=y+1\). Find the value of x*y*z?

A. 1
B. -1/8
C. 1 or 1/8
D. 1 or -(1/8)
E. 1/3

Originally posted by nick1816 on 21 Apr 2019, 22:52.
Last edited by Bunuel on 22 Apr 2019, 00:38, edited 1 time in total.
Renamed the topic.
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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1  [#permalink]

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New post 22 Apr 2019, 04:02
This seems like a formula based question. Could someone post a solution on how to approach it?
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Re: It is given that x^2 + y^2 = z + 1, y^2 +z^2 = x + 1  [#permalink]

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New post 22 Apr 2019, 04:52
3
\(x^2 + y^2\) = z + 1 and \(y^2 + z^2\) = x + 1
subtracting these equations, we get \(x^2 - z^2\)=z-x
or (x+z )(x-z) + (x-z) =0
or (x-z) (x+z+1 ) =0
Hence either x=z or x+z=-1
Similarly you can prove either y=z or y+z=-1
Either way we will get x=y=z

Now we have \(x^2 + y^\)2 = z + 1 or \(x^2 + x^2\) = x + 1
2\(x^2 - x\) - 1=0
x= 1 or -(1/2)

Whenever you see these symmetrical equations in a particular question, mostly you have to either subtract them or add them. Then factorize the resulting equation.
neha283 wrote:
This seems like a formula based question. Could someone post a solution on how to approach it?
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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1  [#permalink]

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New post 23 Apr 2019, 08:13
Can someone please help with this?
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Re: It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1   [#permalink] 23 Apr 2019, 08:13
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It is given that x^2 + y^2 = z + 1, y^2 + z^2 = x + 1 and z^2 +x^2=y+1

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