Solve 1/9a^2+6a+1-1/3a+1diva^2/3a^2+a

$$\frac{ 1 }{ 9 { a }^{ 2 } +6a+1 } - \frac{ 1 }{ 3a+1 } \div \frac{ { a }^{ 2 } }{ 3 { a }^{ 2 } +a }$$

Evaluate

$$-\frac{9a^{2}+5a+1}{a\left(3a+1\right)^{2}}$$

Short Solution Steps

Divide \frac{1}{3a+1} by \frac{a^{2}}{3a^{2}+a} by multiplying \frac{1}{3a+1} by the reciprocal of \frac{a^{2}}{3a^{2}+a}.

$$\frac{1}{9a^{2}+6a+1}-\frac{3a^{2}+a}{\left(3a+1\right)a^{2}}$$

Factor the expressions that are not already factored in \frac{3a^{2}+a}{\left(3a+1\right)a^{2}}.

$$\frac{1}{9a^{2}+6a+1}-\frac{a\left(3a+1\right)}{\left(3a+1\right)a^{2}}$$

Cancel out a\left(3a+1\right) in both numerator and denominator.

$$\frac{1}{9a^{2}+6a+1}-\frac{1}{a}$$

Factor 9a^{2}+6a+1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(3a+1\right)^{2} and a is a\left(3a+1\right)^{2}. Multiply \frac{1}{\left(3a+1\right)^{2}} times \frac{a}{a}. Multiply \frac{1}{a} times \frac{\left(3a+1\right)^{2}}{\left(3a+1\right)^{2}}.

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

Since \frac{a}{a\left(3a+1\right)^{2}} and \frac{\left(3a+1\right)^{2}}{a\left(3a+1\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in a-\left(3a+1\right)^{2}.

$$\frac{a-9a^{2}-6a-1}{a\left(3a+1\right)^{2}}$$

Combine like terms in a-9a^{2}-6a-1.

$$\frac{-5a-9a^{2}-1}{a\left(3a+1\right)^{2}}$$

Expand a\left(3a+1\right)^{2}.

$$\frac{-5a-9a^{2}-1}{9a^{3}+6a^{2}+a}$$

Differentiate w.r.t. a

$$\frac{27a^{3}+21a^{2}+9a+1}{a^{2}\left(3a+1\right)^{3}}$$