Let a,b,c be different nonzero real numbers and x,y,z be three numbers satisfying the system of equations
`(x)/(a)+(y)/(a-1)+(z)/(a+1)=1`
`(x)/(b)+(y)/(b-1)+(z)/(b+1)=1`
`(x)/(c)+(y)/(c-1)+(z)/(c+1)=1`
Then which of the following is correct?

Options

A. `x+y+z=a+b+c`
B. `x=-abc`
C. `z=((1+a)(a+b)(a+c))/(2)`
D. `y=((1-a)(a-b)(a-c))/(2)`

Explanation

To solve the given system of equations, we start with the three equations provided:

1. \(\frac{x}{a} + \frac{y}{a-1} + \frac{z}{a+1} = 1\)
2. \(\frac{x}{b} + \frac{y}{b-1} + \frac{z}{b+1} = 1\)
3. \(\frac{x}{c} + \frac{y}{c-1} + \frac{z}{c+1} = 1\)

### Step 1: Generalize the equations
We can write a generalized equation for any \(t\) (where \(t\) can be \(a\), \(b\), or \(c\)):
\[ \frac{x}{t} + \frac{y}{t-1} + \frac{z}{t+1} = 1 \]

### Step 2: Clear the denominators
Multiply through by \(t(t-1)(t+1)\):
\[ x(t-1)(t+1) + y(t)(t+1) + z(t)(t-1) = t(t-1)(t+1) \]

### Step 3: Expand the equation
Expanding both sides gives:
\[ x(t^2 - 1) + yt(t + 1) + zt(t - 1) = t(t^2 - 1) \]
This simplifies to:
\[ xt^2 - x + yt^2 + yt + zt^2 - zt = t^3 - t \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ (xt^2 + yt^2 + zt^2) + (yt - zt - x) = t^3 - t \]
This can be written as:
\[ ( x + y + z)t^2 + (y - z - x)t + ( -t^3 + t) = 0 \]

### Step 5: Identify the coefficients
This is a cubic equation in \(t\):
\[ t^3 - (x + y + z)t^2 + (y - z - x)t + 0 = 0 \]
Let’s denote the coefficients:
- \(p = x + y + z\)
- \(q = y - z - x\)
- \(r = 0\)

### Step 6: Roots of the cubic equation
Since \(a\), \(b\), and \(c\) are roots of this cubic equation, we can use Vieta's formulas:
1. The sum of the roots \(a + b + c = x + y + z\).
2. The sum of the products of the roots taken two at a time \(ab + ac + bc = - (y - z - x)\).
3. The product of the roots \(abc = 0\).

### Step 7: Analyze the options
From the above, we can conclude:
1. \(x + y + z = a + b + c\) (Correct)
2. \(x = -abc\) (Correct, since \(abc\) is the product of roots)
3. We can find \(y\) and \(z\) using the equations derived from Vieta's formulas.

### Step 8: Solve for \(y\) and \(z\)
From \(x + y + z = a + b + c\):
\[ y + z = a + b + c - x \]
From \(ab + ac + bc = - (y - z - x)\):
\[ y - z = ab + ac + bc + 1 \]

### Step 9: Solve the linear equations for \(y\) and \(z\)
Adding and subtracting these equations will give us the values for \(y\) and \(z\).

### Conclusion
The correct options are:
1. \(x + y + z = a + b + c\)
2. \(x = -abc\)