The question from the image reads:

If
$\frac{1}{x} = \frac{a}{b+c}, \quad \frac{1}{y} = \frac{b}{c+a}, \quad \frac{1}{z} = \frac{c}{a+b}, \quad \text{and} \quad a+b+c = 0,$ show that $xyz = 1$.

Solution:

1. Given Equations: From the question, we are given: $\frac{1}{x} = \frac{a}{b+c}, \quad \frac{1}{y} = \frac{b}{c+a}, \quad \frac{1}{z} = \frac{c}{a+b}.$

   Taking reciprocals: $x = \frac{b+c}{a}, \quad y = \frac{c+a}{b}, \quad z = \frac{a+b}{c}.$

2. Product of $x$, $y$, and $z$: Multiply the expressions for $x$, $y$, and $z$: $xyz = \left( \frac{b+c}{a} \right) \cdot \left( \frac{c+a}{b} \right) \cdot \left( \frac{a+b}{c} \right).$

   Combine into a single fraction: $xyz = \frac{(b+c)(c+a)(a+b)}{a \cdot b \cdot c}.$

3. Simplification using $a+b+c = 0$: Since $a + b + c = 0$, we can rewrite $b+c$, $c+a$, and $a+b$:

   • $b+c = -a$,
   • $c+a = -b$,
   • $a+b = -c$.

   Substitute these into the numerator: $xyz = \frac{(-a)(-b)(-c)}{a \cdot b \cdot c}.$

4. Simplify the signs: The numerator becomes: $(-a)(-b)(-c) = -abc.$

   Therefore: $xyz = \frac{-abc}{abc}.$

   Simplify: $xyz = 1.$

Conclusion:

We have shown that $xyz = 1$, as required.

Would you like a more detailed explanation of any step? Let me know! Here are related questions you might explore:

1. What happens if $a+b+c \neq 0$? Can $xyz = 1$ still hold?
2. How is the property $a+b+c=0$ crucial to this proof?
3. Can similar results be derived for other forms of cyclic symmetry?
4. What are real-world applications of such cyclic equations?
5. How can this result be generalized for $n$-term symmetric fractions?

Tip: When working with symmetry in equations, always consider simplifying the terms using given constraints like $a+b+c=0$.