Question

If x+y+z=0 and x^2+y^2+z^2=7, then x^2y^2+y^2z^2+z^2x^2=a/b, where a and b are positive coprime integers. What is a+b?

Answer 1

Just kidding.. :)

Answer 2

What ? o.O

Answer 3

we have \[
(x+y+z)^2=0\\
\sum x+2\sum xy=0\\
xy+yz+zx=0
\]

Answer 4

similarly,
\[
(xy+yz+zx)^2=0\\
\sum(xy)^2+2\sum(xy)(yz)=0\\
{a\over b}=-2(xy^2z+zyz^2+x^yz)=-2(xyz)(x^2+y^2+z^2)
\]

Answer 5

Wondering how you could get \(\large xy + yz + zx = 0\)
\[\large (x+y+z)^2 = 0 = x^2+y^2+z^2+2xy+2yz+2zx\]
\[\large 0 = 7 + 2xy + 2yz+2zx\]
\[\huge -\frac72= xy+yz+zx\]

Answer 6

doh!
:P

Answer 7

but thats the logic.

Answer 8

This question is not good for my health... I'm getting nauseous :D

Answer 9

lol i got that -7/2 result too :p

Answer 10

Time to toss ideas around...
\[\large (2xy+2yz+2zx)^2=4(x^2y^2+y^2z^2+z^2x^2)+8x^2yz+8xy^2z+8xyz^2\]

Answer 11

bloody hell
\[\ (2xy+2yz+2zx)^2=4(x^2y^2+y^2z^2+z^2x^2)+8x^2yz+8xy^2z+8xyz^2\]

Answer 12

\[
x+y+z=0
\]


\[
x^{2}+y^{2}+z^{2}=7
\]


\[
x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}=\frac{a}{b}\mbox{, a and b positive and coprime}
\]


find \(a+b\)

\[
(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2zx=0
\]


\[
xy+yz+zx=-\frac{7}{2}
\]


Squaring both sides

\[
x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+2x^{2}yz+2xy^{2}z+2xyz^{2}=\frac{49}{4}
\]


\[
x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+2(xyz)(x+y+z)=\frac{49}{4}
\]


since \(x+y+z=0\)

\[
x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}=\frac{49}{4}
\]

Answer 13

49/4 can't be simplified any further and 49 and 4 are coprime so a + b = 53

Answer 14

I cannot believe I missed that , @Meepi :)

Answer 15

hey .. how about a medal to get you guys started???

Answer 16

:D

Answer 17

Now, where is @vikrantg4 ?

Answer 18

I was thinking the same but I thought that I would not make it to the final answer...

Answer 19

LoL meepi rocked !