The solutions of the equations, where `a+b+c ne 0`
`(b+c) (y+z) -ax =b-c`
`(c+a) (z+x)-by =c-a`
`(a+b)(x+y)-cz =a-b` are ….

Answer

To solve the given equations, we will follow a systematic approach. The equations provided are:

1. \((b+c)(y+z) - ax = b - c\)
2. \((c+a)(z+x) - by = c - a\)
3. \((a+b)(x+y) - cz = a - b\)

We need to find the values of \(x\), \(y\), and \(z\).

### Step 1: Rearranging the Equations
We will rearrange each equation to isolate the terms involving \(x\), \(y\), and \(z\).

1. From the first equation:
\[ (b+c)(y+z) = ax + b - c \]
This can be rewritten as:
\[ y + z = \frac{ax + b - c}{b+c} \]

2. From the second equation:
\[ (c+a)(z+x) = by + c - a \]
This can be rewritten as:
\[ z + x = \frac{by + c - a}{c+a} \]

3. From the third equation:
\[ (a+b)(x+y) = cz + a - b \]
This can be rewritten as:
\[ x + y = \frac{cz + a - b}{a+b} \]

### Step 2: Adding the Equations
Now, we will add all three rearranged equations together:
\[ (y + z) + (z + x) + (x + y) = \frac{ax + b - c}{b+c} + \frac{by + c - a}{c+a} + \frac{cz + a - b}{a+b} \]

This simplifies to:
\[ 2(x + y + z) = \frac{ax + b - c}{b+c} + \frac{by + c - a}{c+a} + \frac{cz + a - b}{a+b} \]

### Step 3: Setting Up the Equation
Since \(a + b + c \neq 0\), we can factor out \(a + b + c\) from the right-hand side. This leads us to:
\[ x + y + z = 0 \]

### Step 4: Substituting Values
Now we will substitute \(z = -x - y\) into the original equations to find \(x\) and \(y\).

1. Substitute \(z\) in the first equation:
\[ (b+c)(y - x - y) - ax = b - c \]
Simplifying gives:
\[ (b+c)(-x) - ax = b - c \]
Rearranging gives:
\[ -(b+c + a)x = b - c \]

2. From the second equation:
\[ (c+a)(-x) - by = c - a \]
Rearranging gives:
\[ -(c+a)x - by = c - a \]

3. From the third equation:
\[ (a+b)(x + y) - c(-x - y) = a - b \]
Rearranging gives:
\[ (a+b + c)(x + y) = a - b \]

### Step 5: Solving for \(x\), \(y\), and \(z\)
Now we can solve for \(x\), \(y\), and \(z\) using the equations derived from the substitutions.

1. From the first equation, we can express \(x\):
\[ x = \frac{b - c}{a + b + c} \]

2. Substitute \(x\) into the second equation to find \(y\):
\[ y = \frac{c - a}{a + b + c} \]

3. Finally, substitute \(x\) and \(y\) into the expression for \(z\):
\[ z = -\left(\frac{b - c}{a + b + c} + \frac{c - a}{a + b + c}\right) = \frac{a - b}{a + b + c} \]

### Final Solutions
Thus, the solutions for \(x\), \(y\), and \(z\) are:
\[ x = \frac{b - c}{a + b + c}, \quad y = \frac{c - a}{a + b + c}, \quad z = \frac{a - b}{a + b + c} \]