maths question on simultaneous equations

Question

Solve the following system of simultaneous equations,
4xyz+xy+yz+zx=-1
X(y+z)+y(z+x)+z(x+y)=6
X^2+y^2+z^2+6x+6y+6z=-15

Answer 1

 

One solution is (-1,-1,-1) found numerically

 

1st equation:
4xyz + xy + yz + zx = -1


2nd equation:
xy + xz + yz + xy + xz + yz = 6
2xy + 2xz + 2yz = 6
xy + xz + yz = 3

3rd equation:
x^2 + y^2 + z^2 + 6x + 6y + 6z = -15

(x^2 + 6x ) + (y^2 + 6y ) + (Z^2 + 6z ) = -15

(x^2 + 6x + 9) + (y^2 + 6y + 9) + (z^2 + 6z + 9) = -15 + 27 <--- completes the square

(x+3)^2 + (y+3)^2 + (z+3)^2 = 12



Substitutes the second equation into the first:

4xyz + 3 = -1


4xyz = -4

xyz = -1


So per the third equation we have a spehere at (-3,-3,-3) with radius 2*sqrt(3)

Meanwhile xyz = -1 is a hyperboloid



We are interested in finding where the sphere and the hyperboloid shall
intersect

(x+3)^2 + (y+3)^2+(z+3)^2 - xyz = 12 + 1 = 13

F(x,y,z) = (x+3)^2 + (y+3)^2+(z+3)^2 - xyz - 13 = 0


One solution is (-1,-1,-1)