If a = frac{x}{y-z},b = frac{y}{z-x},and c = | vedclass

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(B) Given $a = \frac{x}{y-z}$,$b = \frac{y}{z-x}$,$c = \frac{z}{x-y}$.Let us test with values $x=1, y=2, z=4$ (ensuring denominators are non-zero).$a

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-$1$

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(B) Given $a = \frac{x}{y-z}$,$b = \frac{y}{z-x}$,$c = \frac{z}{x-y}$.Let us test with values $x=1, y=2, z=4$ (ensuring denominators are non-zero).$a = \frac{1}{2-4} = -\frac{1}{2}$$b = \frac{2}{4-1} = \frac{2}{3}$$c = \frac{4}{1-2} = -4$Now calculate $ab + bc + ca + abc$:$ab = (-\frac{1}{2})(\frac{2}{3}) = -\frac{1}{3}$$bc = (\frac{2}{3})(-4) = -\frac{8}{3}$$ca = (-4)(-\frac{1}{2}) = 2$$abc = (-\frac{1}{2})(\frac{2}{3})(-4) = \frac{4}{3}$Sum $= -\frac{1}{3} - \frac{8}{3} + 2 + \frac{4}{3} = \frac{-1-8+4}{3} + 2 = -\frac{5}{3} + 2 = \frac{1}{3}$.Note: The original expression $ab+bc+ca$ evaluates to $-1$ for these specific variables.

If $a = \frac{x}{y-z}$,$b = \frac{y}{z-x}$,and $c = \frac{z}{x-y}$,where $x, y$,and $z$ are distinct such that $x-y, y-z, z-x \neq 0$,then what is the value of $ab + bc + ca + abc$?

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