Solve =(x-y)^3+(y-z)^3+(z-x)^3/9(x-y)(y-z)(z-x)

$$= \frac { ( x - y ) ^ { 3 } + ( y - z ) ^ { 3 } + ( z - x ) ^ { 3 } } { 9 ( x - y ) ( y - z ) ( z - x ) }$$

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{3\left(x-y\right)\left(x-z\right)\left(-y+z\right)}{9\left(x-y\right)\left(y-z\right)\left(-x+z\right)}$$

Extract the negative sign in x-z. Extract the negative sign in -y+z.

$$\frac{-\left(-1\right)\times 3\left(x-y\right)\left(y-z\right)\left(-x+z\right)}{9\left(x-y\right)\left(y-z\right)\left(-x+z\right)}$$

Cancel out 3\left(x-y\right)\left(y-z\right)\left(-x+z\right) in both numerator and denominator.

$$\frac{-\left(-1\right)}{3}$$

Multiply -1 and -1 to get 1.

$$\frac{1}{3}$$

Factor

$$\frac{1}{3} = 0.3333333333333333$$