Question

If \(x^2 = y + z, y^2 = z + x\) and \(z^2 = x + y\) then the value of \(\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1}\) is

• A. \(-1\)
• B. \(1\)
• C. \(2\)
• D. \(4\)

Hint:

For symmetric variables, try assuming \(x = y = z\).
Then \(x^2 = x + x \implies x^2 = 2x \implies x = 2\).
Substituting $x=2$: $\frac{1}{2+1} + \frac{1}{2+1} + \frac{1}{2+1} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a symmetric algebraic identity problem. We can transform each term to have the same common denominator.
Step 2: Detailed Explanation:
1. Consider the first given equation: \(x^2 = y + z\).
Add \(x\) to both sides:
\[ x^2 + x = x + y + z \] \[ x(x + 1) = x + y + z \implies x + 1 = \frac{x+y+z}{x} \] Take the reciprocal:
\[ \frac{1}{x+1} = \frac{x}{x+y+z} \] 2. Similarly, for \(y^2 = z + x\), adding \(y\) to both sides gives:
\[ y(y + 1) = x + y + z \implies \frac{1}{y+1} = \frac{y}{x+y+z} \] 3. For \(z^2 = x + y\), adding \(z\) to both sides gives:
\[ z(z + 1) = x + y + z \implies \frac{1}{z+1} = \frac{z}{x+y+z} \] 4. Summing all three derived terms:
\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = \frac{x}{x+y+z} + \frac{y}{x+y+z} + \frac{z}{x+y+z} \] \[ = \frac{x + y + z}{x + y + z} = 1 \] Step 3: Final Answer:
The value of the expression is \(1\).