Solve 2a-b-c/(a-b)(a-c)+2b-c-a/(b-c)(b-a)+2c-b-a/(c-b)(c-a)

$$\frac { 2 a - b - c } { ( a - b ) ( a - c ) } + \frac { 2 b - c - a } { ( b - c ) ( b - a ) } + \frac { 2 c - b - a } { ( c - b ) ( c - a ) }$$

Evaluate

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Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(a-b\right)\left(a-c\right) and \left(b-c\right)\left(b-a\right) is \left(a-b\right)\left(a-c\right)\left(-b+c\right). Multiply \frac{2a-b-c}{\left(a-b\right)\left(a-c\right)} times \frac{-b+c}{-b+c}. Multiply \frac{2b-c-a}{\left(b-c\right)\left(b-a\right)} times \frac{a-c}{a-c}.

$$\frac{\left(2a-b-c\right)\left(-b+c\right)}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)}+\frac{\left(2b-c-a\right)\left(a-c\right)}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)}+\frac{2c-b-a}{\left(c-b\right)\left(c-a\right)}$$

Since \frac{\left(2a-b-c\right)\left(-b+c\right)}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)} and \frac{\left(2b-c-a\right)\left(a-c\right)}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(2a-b-c\right)\left(-b+c\right)+\left(2b-c-a\right)\left(a-c\right)}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)}+\frac{2c-b-a}{\left(c-b\right)\left(c-a\right)}$$

Do the multiplications in \left(2a-b-c\right)\left(-b+c\right)+\left(2b-c-a\right)\left(a-c\right).

$$\frac{-2ab+2ac+b^{2}-bc+cb-c^{2}+2ba-2bc-ca+c^{2}-a^{2}+ac}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)}+\frac{2c-b-a}{\left(c-b\right)\left(c-a\right)}$$

Combine like terms in -2ab+2ac+b^{2}-bc+cb-c^{2}+2ba-2bc-ca+c^{2}-a^{2}+ac.

$$\frac{2ac+b^{2}-2bc-a^{2}}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)}+\frac{2c-b-a}{\left(c-b\right)\left(c-a\right)}$$

Factor the expressions that are not already factored in \frac{2ac+b^{2}-2bc-a^{2}}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)}.

$$\frac{\left(-a+b\right)\left(a+b-2c\right)}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)}+\frac{2c-b-a}{\left(c-b\right)\left(c-a\right)}$$

Extract the negative sign in -a+b.

$$\frac{-\left(a-b\right)\left(a+b-2c\right)}{\left(a-b\right)\left(a-c\right)\left(-b+c\right)}+\frac{2c-b-a}{\left(c-b\right)\left(c-a\right)}$$

Cancel out a-b in both numerator and denominator.

$$\frac{-\left(a+b-2c\right)}{\left(a-c\right)\left(-b+c\right)}+\frac{2c-b-a}{\left(c-b\right)\left(c-a\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(a-c\right)\left(-b+c\right) and \left(c-b\right)\left(c-a\right) is \left(-a+c\right)\left(-b+c\right). Multiply \frac{-\left(a+b-2c\right)}{\left(a-c\right)\left(-b+c\right)} times \frac{-1}{-1}.

$$\frac{-\left(-1\right)\left(a+b-2c\right)}{\left(-a+c\right)\left(-b+c\right)}+\frac{2c-b-a}{\left(-a+c\right)\left(-b+c\right)}$$

Since \frac{-\left(-1\right)\left(a+b-2c\right)}{\left(-a+c\right)\left(-b+c\right)} and \frac{2c-b-a}{\left(-a+c\right)\left(-b+c\right)} have the same denominator, add them by adding their numerators.

$$\frac{-\left(-1\right)\left(a+b-2c\right)+2c-b-a}{\left(-a+c\right)\left(-b+c\right)}$$

Do the multiplications in -\left(-1\right)\left(a+b-2c\right)+2c-b-a.

$$\frac{a+b-2c+2c-b-a}{\left(-a+c\right)\left(-b+c\right)}$$

Combine like terms in a+b-2c+2c-b-a.

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

Zero divided by any non-zero term gives zero.

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Factor

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