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Question

Solve the following equations :
y2+z2(y+z)x=12, x2+z2(x+z)y=8,
x2+y2(x+y)z=2.

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Solution

We rewrite the equations as
(y2zx)+(z2xy)=12
(x2yz)+(z2xy)=8
(x2yz)+(y2zx)=2
Adding (1),(2) and (3), we get
2(y2zx)+2(z2xy)+2(x2yz)=22
or (y2zx)+(z2xy)+(x2yz)=11
From (1) and (4),
x2yz=1
Similarly from (2) and (4),
y2zx=3
and from (3) and (4),
z2xy=9
Multiplying (5),(6),(7) by y,z,x respectively and adding, we get
y+3z+9x=0 or 9xy+3z=0
Again multiplying these equations by z,x and y and adding, we get
z+3x+9y=0
or 3x+9yz=0
Solving (8) and (9) by cross-multiplication, we get
x127=y9+9=z81+3
or x26=y18=z84
or x13=y9=z42=k,say
Substituting these values in (5), we get
169k2378k2=1
or 209k2=1 or k=±1(209)
Hence x=±13(209),y=±9(209),z=±42(209).

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