If `(x^(2))/(by+cz)=(y^(2))/(cz+ax)=(z^(2))/(ax+by)=1`, then value of `(a)/(a+x) +(b)/(b+y) +(c )/(c+z)` is

Options

A. `-1`
B. 2
C. 1
D. `-2`

Answer

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

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

This implies:

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

Now, we will express \( x^2, y^2, z^2 \) in terms of \( a, b, c, x, y, z \):

From the first equation:
\[ x^2 = by + cz \]
Adding \( ax \) to both sides:
\[ x^2 + ax = by + cz + ax \]
Factoring out \( x \):
\[ x(x + a) = by + cz + ax \]

From the second equation:
\[ y^2 = cz + ax \]
Adding \( by \) to both sides:
\[ y^2 + by = cz + ax + by \]
Factoring out \( y \):
\[ y(y + b) = cz + ax + by \]

From the third equation:
\[ z^2 = ax + by \]
Adding \( cz \) to both sides:
\[ z^2 + cz = ax + by + cz \]
Factoring out \( z \):
\[ z(z + c) = ax + by + cz \]

Now we need to find the value of:
\[ \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} \]

Substituting the expressions we derived:
\[ \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} = \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} \]

Now let's rewrite each term:
1. For \( \frac{a}{a+x} \):
\[ \frac{a}{a+x} = \frac{a}{a + \frac{x^2}{by + cz}} = \frac{a}{a + \frac{by + cz}{by + cz}} = \frac{a(by + cz)}{(a + by + cz)} \]

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

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

Now, adding these fractions together, we notice that the denominators are the same:
\[ \frac{a + b + c}{a + b + c} = 1 \]

Thus, the value of \( \frac{a}{a+x} + \frac{b}{b+y} + \frac{c}{c+z} \) simplifies to:

\[ \boxed{1} \]