Solve 6a^2-6b^2/b-a3a^+3b^/{a^2+2ab+{b}^2}

$$\frac{ 6 { a }^{ 2 } -6 { b }^{ 2 } }{ b-a } \frac{ 3 { a }^{ } +3 { b }^{ } }{ { a }^{ 2 } +2ab+ { b }^{ 2 } }$$

Evaluate

$$-18$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{6a^{2}-6b^{2}}{b-a}.

$$\frac{6\left(a+b\right)\left(a-b\right)}{-a+b}\times \frac{3a^{1}+3b^{1}}{a^{2}+2ab+b^{2}}$$

Extract the negative sign in a-b.

$$\frac{-6\left(a+b\right)\left(-a+b\right)}{-a+b}\times \frac{3a^{1}+3b^{1}}{a^{2}+2ab+b^{2}}$$

Cancel out -a+b in both numerator and denominator.

$$-6\left(a+b\right)\times \frac{3a^{1}+3b^{1}}{a^{2}+2ab+b^{2}}$$

Expand the expression.

$$\left(-6a-6b\right)\times \frac{3a^{1}+3b^{1}}{a^{2}+2ab+b^{2}}$$

Calculate a to the power of 1 and get a.

$$\left(-6a-6b\right)\times \frac{3a+3b^{1}}{a^{2}+2ab+b^{2}}$$

Calculate b to the power of 1 and get b.

$$\left(-6a-6b\right)\times \frac{3a+3b}{a^{2}+2ab+b^{2}}$$

Factor the expressions that are not already factored in \frac{3a+3b}{a^{2}+2ab+b^{2}}.

$$\left(-6a-6b\right)\times \frac{3\left(a+b\right)}{\left(a+b\right)^{2}}$$

Cancel out a+b in both numerator and denominator.

$$\left(-6a-6b\right)\times \frac{3}{a+b}$$

Express \left(-6a-6b\right)\times \frac{3}{a+b} as a single fraction.

$$\frac{\left(-6a-6b\right)\times 3}{a+b}$$

Factor the expressions that are not already factored.

$$\frac{3\times 6\left(-a-b\right)}{a+b}$$

Extract the negative sign in -a-b.

$$\frac{-3\times 6\left(a+b\right)}{a+b}$$

Cancel out a+b in both numerator and denominator.

$$-3\times 6$$

Multiply -3 and 6 to get -18.

$$-18$$

Factor

$$-18$$