(x+y)(y+z)=5(x+y)(y+z)

(x+z)(y+z)=7(y+z)(x+z)
(x+y)(x+z)=6(z+x)(x+y)
find xyz
show work plz

(x+y)(y+z)=5(x+y)(y+z)

one cow = 5 cows
Please check for typos

clearly x = -y = -z

y = -z

so, they are all zero.

To find the value of xyz, we can start by simplifying each of the given equations.

1) (x+y)(y+z) = 5(x+y)(y+z)
Expanding the left side and right side:
xy + xz + y^2 + yz = 5xy + 5xz + 5y^2 + 5yz

Rearranging the terms:
xy - 5xy + xz - 5xz + y^2 - 5y^2 + yz - 5yz = 0

Combining like terms:
-4xy - 4xz - 4y^2 - 4yz = 0

Dividing both sides by -4:
xy + xz + y^2 + yz = 0

2) (x+z)(y+z) = 7(y+z)(x+z)
Expanding the left side and right side:
xy + xz + yz + z^2 = 7xy + 7xz + 7yz + 7z^2

Rearranging the terms:
xy - 7xy + xz - 7xz + yz - 7yz + z^2 - 7z^2 = 0

Combining like terms:
-6xy - 6xz - 6yz - 6z^2 = 0

Dividing both sides by -6:
xy + xz + yz + z^2 = 0

3) (x+y)(x+z) = 6(z+x)(x+y)
Expanding the left side and right side:
x^2 + xy + xz + yz = 6zx + 6xz + 6zy + 6xy

Rearranging the terms:
x^2 - 6zx + xy - 6xy + xz - 6xz + yz - 6zy = 0

Combining like terms:
x^2 - 5zx - 5xy - 5xz - 5zy + yz = 0

Now, combining all three equations together:
xy + xz + y^2 + yz = 0
xy + xz + yz + z^2 = 0
x^2 - 5zx - 5xy - 5xz - 5zy + yz = 0

From the first equation, we can rewrite y^2 as -xy - xz - yz:
xy + xz - xy - xz - yz + yz = 0
0 = 0

This implies that the first equation does not provide any additional information.

Similarly, from the second equation, we can rewrite z^2 as -xy - xz - yz:
xy + xz + yz - xy - xz - yz = 0
0 = 0

Again, this implies that the second equation does not provide any additional information.

Now, let's substitute the value of z^2 from the second equation into the third equation:
x^2 - 5zx - 5xy - 5xz - 5zy + xy + xz + yz = 0
x^2 - 5zx - 5xy - 5xz - 5zy + xy + xz + yz = 0

Simplifying:
x^2 - 4xy - 4xz - 5zx - 5zy = 0

Rearranging:
x^2 - 4xy - 4xz - 5zx - 5zy = 0
Factorizing out x:
x(x - 4y - 4z - 5z) = 0

For this equation to hold true, either x = 0 or x - 4y - 4z - 5z = 0.

If we assume x = 0, then from the first equation:
0 + 0 + y^2 + yz = 0
y(y + z) = 0

Here, either y = 0 or y + z = 0.

If y = 0, then from the second equation:
0 + 0 + 0 + z^2 = 0
z^2 = 0

Therefore, z must also be 0.

So, when x = 0, y = 0, and z = 0, the equations are satisfied.

Hence, the value of xyz is 0.