Word Count : 563

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We are asked to solve the system: $$ \frac{dx}{y-z} = \frac{dy}{z-x} = \frac{dz}{x-y}. $$ Let us denote the common ratio by $k$, so $$ dx = k(y-z), \quad dy = k(z-x), \quad dz = k(x-y). $$ Step 1: Sum of differentials Consider $dx + dy + dz$: $$ dx + dy + dz = k[(y-z) + (z-x) + (x-y)] = k \cdot 0 = 0 \implies dx + dy + dz = 0. $$ Integrating, we get $$ x + y + z = C_1, \quad C_1 \text{ is a constant.} $$ Step 2: Consider differences Take $dx - dy$: $$ dx - dy = k[(y-z) - (z-x)] = k[y-z-z+x] = k(x+y-2z) $$ Similarly, $dy - dz = k[(z-x)-(x-y)] = k(-2x+y+z)$, and $dz - dx = k[(x-y)-(y-z)] = k(x-2y+z)$. Step 3: Introduce substitution using sum Since $x+y+z = C_1$, let _____ ______ _________ ____ __________ _________.
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