`(x^(2))/(by + cz) = (y^(2))/(ax + cz)= (z^(2))/(ax + by)=1` find
`(a)/(x+a) + (b)/(y+b)+ (c )/(z+c )`= ?

Answer

To solve the problem, we start with the given equations:

\[ \frac{x^2}{by + cz} = \frac{y^2}{ax + cz} = \frac{z^2}{ax + by} = 1 \]

From this, we can derive three separate equations:

1. \(\frac{x^2}{by + cz} = 1\)
2. \(\frac{y^2}{ax + cz} = 1\)
3. \(\frac{z^2}{ax + by} = 1\)

### Step 1: Rearranging the equations

From the first equation, we can rearrange it to find \(by + cz\):

\[ x^2 = by + cz \implies by + cz = x^2 \]

From the second equation, we can find \(ax + cz\):

\[ y^2 = ax + cz \implies ax + cz = y^2 \]

From the third equation, we can find \(ax + by\):

\[ z^2 = ax + by \implies ax + by = z^2 \]

### Step 2: Substituting the values

Now we have:

1. \(by + cz = x^2\)
2. \(ax + cz = y^2\)
3. \(ax + by = z^2\)

### Step 3: Finding the expression

We need to find:

\[ \frac{a}{x + a} + \frac{b}{y + b} + \frac{c}{z + c} \]

To simplify this, we can express each term in the form of the equations we derived.

### Step 4: Simplifying each term

For the first term:

\[ \frac{a}{x + a} = \frac{a}{x + a} \cdot \frac{x^2}{x^2} = \frac{ax^2}{x^2(x + a)} = \frac{ax^2}{x^2 + ax^2} \]

For the second term:

\[ \frac{b}{y + b} = \frac{b}{y + b} \cdot \frac{y^2}{y^2} = \frac{by^2}{y^2(y + b)} = \frac{by^2}{y^2 + by^2} \]

For the third term:

\[ \frac{c}{z + c} = \frac{c}{z + c} \cdot \frac{z^2}{z^2} = \frac{cz^2}{z^2(z + c)} = \frac{cz^2}{z^2 + cz^2} \]

### Step 5: Combining the terms

Now, we can combine these fractions:

\[ \frac{ax^2}{x^2 + ax^2} + \frac{by^2}{y^2 + by^2} + \frac{cz^2}{z^2 + cz^2} \]

### Step 6: Final simplification

Since \(by + cz = x^2\), \(ax + cz = y^2\), and \(ax + by = z^2\), we can see that:

\[ \frac{ax^2}{x^2 + ax^2} + \frac{by^2}{y^2 + by^2} + \frac{cz^2}{z^2 + cz^2} = 1 \]

Thus, the final answer is:

\[ \frac{a}{x + a} + \frac{b}{y + b} + \frac{c}{z + c} = 1 \]