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

Answer

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

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

This implies:

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

We need to find:

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

### Step 1: Express \( x, y, z \) in terms of \( a, b, c \)

From the equations, we can express \( by + cz \), \( ax + cz \), and \( ax + by \) in terms of \( x, y, z \):

- From \( x^2 = by + cz \), we can write:
\[ by + cz = x^2 \]

- From \( y^2 = ax + cz \), we can write:
\[ ax + cz = y^2 \]

- From \( z^2 = ax + by \), we can write:
\[ ax + by = z^2 \]

### Step 2: Substitute these expressions into the desired sum

Now we substitute these expressions into the sum we want to find:

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

### Step 3: Simplify each term

We can rewrite each term:

1. For \( \frac{x}{x+a} \):
\[ \frac{x}{x+a} = \frac{x}{x + \frac{x^2 - (by + cz)}{x}} = \frac{x^2}{x^2 + ax} \]

2. For \( \frac{y}{y+b} \):
\[ \frac{y}{y+b} = \frac{y^2}{y^2 + by} \]

3. For \( \frac{z}{z+c} \):
\[ \frac{z}{z+c} = \frac{z^2}{z^2 + cz} \]

### Step 4: Combine the fractions

Now we can combine these fractions:

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

### Step 5: Substitute the values

Substituting the values from the earlier equations:

\[ = \frac{by + cz}{by + cz + ax} + \frac{ax + cz}{ax + cz + by} + \frac{ax + by}{ax + by + cz} \]

### Step 6: Find a common denominator and simplify

The common denominator will be:

\[ (by + cz + ax)(ax + cz + by)(ax + by + cz) \]

After substituting and simplifying, we can find that:

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

### Final Answer

Thus, the final answer is:

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