Question:

Given the definitions of $a, b,$ and $c$: $$a = \frac{x}{y-z}, \quad b = \frac{y}{z-x}, \quad c = \frac{z}{x-y}$$ We aim to find the value of $ab + bc + ca$. Step 1: Express $ab + bc + ca$ with a common denominator. First, calculate the individual products: $$ab = \frac{x}{y-z} \cdot \frac{y}{z-x} = \frac{xy}{(y-z)(z-x)}$$ $$bc = \frac{y}{z-x} \cdot \frac{z}{x-y} = \frac{yz}{(z-x)(x-y)}$$ $$ca = \frac{z}{x-y} \cdot \frac{x}{y-z} = \frac{zx}{(x-y)(y-z)}$$ The common denominator for these terms is $(y-z)(z-x)(x-y)$. $$ab + bc + ca = \frac{xy(x-y)}{(y-z)(z-x)(x-y)} + \frac{yz(y-z)}{(y-z)(z-x)(x-y)} + \frac{zx(z-x)}{(y-z)(z-x)(x-y)}$$ $$ab + bc + ca = \frac{xy(x-y) + yz(y-z) + zx(z-x)}{(y-z)(z-x)(x-y)}$$ Step 2: Simplify the numerator using algebraic factorization. Let the numerator be $N$: $$N = xy(x-y) + yz(y-z) + zx(z-x)$$ Expand the terms: $$N = (x^2y - xy^2) + (y^2z - yz^2) + (z^2x - zx^2)$$ This expression is a known cyclic identity, which factors as: $$N = -(x-y)(y-z)(z-x)$$ To verify this factorization, expand the right side: $$-(x-y)(y-z)(z-x) = -(x-y)(yz - xy - z^2 + zx)$$ $$= -(x(yz - xy - z^2 + zx) - y(yz - xy - z^2 + zx))$$ $$= -(xyz - x^2y - xz^2 + x^2z - y^2z + xy^2 + yz^2 - xyz)$$ $$= -(-x^2y - xz^2 + x^2z - y^2z + xy^2 + yz^2)$$ $$= x^2y + xz^2 - x^2z + y^2z - xy^2 - yz^2$$ This matches the expanded form of $N$. Step 3: Substitute the simplified numerator and evaluate the expression. Substitute $N = -(x-y)(y-z)(z-x)$ back into the expression for $ab+bc+ca$: $$ab + bc + ca = \frac{-(x-y)(y-z)(z-x)}{(y-z)(z-x)(x-y)}$$ For $a, b, c$ to be defined, the denominators $y-z, z-x, x-y$ must be non-zero, which implies $x \neq y$, $y \neq z$, and $z \neq x$. Under this condition, we can cancel the common factors in the numerator and denominator: $$ab + bc + ca = -1$$