Solve a/(c-a)(a-b)+b/(a-b)(b-c)+c/(b-c)(c-a)

$$\frac { a } { ( c - a ) ( a - b ) } + \frac { b } { ( a - b ) ( b - c ) } + \frac { c } { ( b - c ) ( c - a ) }$$

Evaluate

$$0$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(c-a\right)\left(a-b\right) and \left(a-b\right)\left(b-c\right) is \left(a-b\right)\left(b-c\right)\left(-a+c\right). Multiply \frac{a}{\left(c-a\right)\left(a-b\right)} times \frac{b-c}{b-c}. Multiply \frac{b}{\left(a-b\right)\left(b-c\right)} times \frac{-a+c}{-a+c}.

$$\frac{a\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)}+\frac{b\left(-a+c\right)}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)}+\frac{c}{\left(b-c\right)\left(c-a\right)}$$

Since \frac{a\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)} and \frac{b\left(-a+c\right)}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)} have the same denominator, add them by adding their numerators.

$$\frac{a\left(b-c\right)+b\left(-a+c\right)}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)}+\frac{c}{\left(b-c\right)\left(c-a\right)}$$

Do the multiplications in a\left(b-c\right)+b\left(-a+c\right).

$$\frac{ab-ac-ba+bc}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)}+\frac{c}{\left(b-c\right)\left(c-a\right)}$$

Combine like terms in ab-ac-ba+bc.

$$\frac{-ac+bc}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)}+\frac{c}{\left(b-c\right)\left(c-a\right)}$$

Factor the expressions that are not already factored in \frac{-ac+bc}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)}.

$$\frac{c\left(-a+b\right)}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)}+\frac{c}{\left(b-c\right)\left(c-a\right)}$$

Extract the negative sign in -a+b.

$$\frac{-c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(-a+c\right)}+\frac{c}{\left(b-c\right)\left(c-a\right)}$$

Cancel out a-b in both numerator and denominator.

$$\frac{-c}{\left(b-c\right)\left(-a+c\right)}+\frac{c}{\left(b-c\right)\left(c-a\right)}$$

Since \frac{-c}{\left(b-c\right)\left(-a+c\right)} and \frac{c}{\left(b-c\right)\left(c-a\right)} have the same denominator, add them by adding their numerators.

$$\frac{-c+c}{\left(b-c\right)\left(-a+c\right)}$$

Combine like terms in -c+c.

$$\frac{0}{\left(b-c\right)\left(-a+c\right)}$$

Zero divided by any non-zero term gives zero.

$$0$$

Factor

$$0$$