Solve a-b/3a+3bdiva^2-b^2/a^2+8a+16

$$\frac { a - b } { 3 a + 3 b } \div \frac { a ^ { 2 } - b ^ { 2 } } { a ^ { 2 } + 8 a + 16 }$$

Evaluate

$$\frac{\left(\frac{a+4}{a+b}\right)^{2}}{3}$$

Short Solution Steps

Divide \frac{a-b}{3a+3b} by \frac{a^{2}-b^{2}}{a^{2}+8a+16} by multiplying \frac{a-b}{3a+3b} by the reciprocal of \frac{a^{2}-b^{2}}{a^{2}+8a+16}.

$$\frac{\left(a-b\right)\left(a^{2}+8a+16\right)}{\left(3a+3b\right)\left(a^{2}-b^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(a-b\right)\left(a+4\right)^{2}}{3\left(a-b\right)\left(a+b\right)^{2}}$$

Cancel out a-b in both numerator and denominator.

$$\frac{\left(a+4\right)^{2}}{3\left(a+b\right)^{2}}$$

Expand the expression.

$$\frac{a^{2}+8a+16}{3a^{2}+6ab+3b^{2}}$$

Expand

$$\frac{a^{2}+8a+16}{3\left(a+b\right)^{2}}$$

Short Solution Steps

Divide \frac{a-b}{3a+3b} by \frac{a^{2}-b^{2}}{a^{2}+8a+16} by multiplying \frac{a-b}{3a+3b} by the reciprocal of \frac{a^{2}-b^{2}}{a^{2}+8a+16}.

$$\frac{\left(a-b\right)\left(a^{2}+8a+16\right)}{\left(3a+3b\right)\left(a^{2}-b^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(a-b\right)\left(a+4\right)^{2}}{3\left(a-b\right)\left(a+b\right)^{2}}$$

Cancel out a-b in both numerator and denominator.

$$\frac{\left(a+4\right)^{2}}{3\left(a+b\right)^{2}}$$

Expand the expression.

$$\frac{a^{2}+8a+16}{3a^{2}+6ab+3b^{2}}$$