Solve 1/b-3+3b^2/27-b^3+3/b^2+3b+9

$$\frac { 1 } { b - 3 } + \frac { 3 b ^ { 2 } } { 27 - b ^ { 3 } } + \frac { 3 } { b ^ { 2 } + 3 b + 9 }$$

Evaluate

$$-\frac{2b}{b^{2}+3b+9}$$

Short Solution Steps

Factor 27-b^{3}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of b-3 and \left(b-3\right)\left(-b^{2}-3b-9\right) is \left(b-3\right)\left(-b^{2}-3b-9\right). Multiply \frac{1}{b-3} times \frac{-b^{2}-3b-9}{-b^{2}-3b-9}.

$$\frac{-b^{2}-3b-9}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3b^{2}}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9}$$

Since \frac{-b^{2}-3b-9}{\left(b-3\right)\left(-b^{2}-3b-9\right)} and \frac{3b^{2}}{\left(b-3\right)\left(-b^{2}-3b-9\right)} have the same denominator, add them by adding their numerators.

$$\frac{-b^{2}-3b-9+3b^{2}}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9}$$

Combine like terms in -b^{2}-3b-9+3b^{2}.

$$\frac{2b^{2}-3b-9}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9}$$

Factor the expressions that are not already factored in \frac{2b^{2}-3b-9}{\left(b-3\right)\left(-b^{2}-3b-9\right)}.

$$\frac{\left(b-3\right)\left(2b+3\right)}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9}$$

Cancel out b-3 in both numerator and denominator.

$$\frac{2b+3}{-b^{2}-3b-9}+\frac{3}{b^{2}+3b+9}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of -b^{2}-3b-9 and b^{2}+3b+9 is b^{2}+3b+9. Multiply \frac{2b+3}{-b^{2}-3b-9} times \frac{-1}{-1}.

$$\frac{-\left(2b+3\right)}{b^{2}+3b+9}+\frac{3}{b^{2}+3b+9}$$

Since \frac{-\left(2b+3\right)}{b^{2}+3b+9} and \frac{3}{b^{2}+3b+9} have the same denominator, add them by adding their numerators.

$$\frac{-\left(2b+3\right)+3}{b^{2}+3b+9}$$

Do the multiplications in -\left(2b+3\right)+3.

$$\frac{-2b-3+3}{b^{2}+3b+9}$$

Combine like terms in -2b-3+3.

$$\frac{-2b}{b^{2}+3b+9}$$

Differentiate w.r.t. b

$$\frac{2\left(b^{2}-9\right)}{\left(b^{2}+3b+9\right)^{2}}$$

Steps Using Derivative Rule for Quotient

Factor 27-b^{3}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{1}{b-3}+\frac{3b^{2}}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of b-3 and \left(b-3\right)\left(-b^{2}-3b-9\right) is \left(b-3\right)\left(-b^{2}-3b-9\right). Multiply \frac{1}{b-3} times \frac{-b^{2}-3b-9}{-b^{2}-3b-9}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{-b^{2}-3b-9}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3b^{2}}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9})$$

Since \frac{-b^{2}-3b-9}{\left(b-3\right)\left(-b^{2}-3b-9\right)} and \frac{3b^{2}}{\left(b-3\right)\left(-b^{2}-3b-9\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{-b^{2}-3b-9+3b^{2}}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9})$$

Combine like terms in -b^{2}-3b-9+3b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{2b^{2}-3b-9}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9})$$

Factor the expressions that are not already factored in \frac{2b^{2}-3b-9}{\left(b-3\right)\left(-b^{2}-3b-9\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{\left(b-3\right)\left(2b+3\right)}{\left(b-3\right)\left(-b^{2}-3b-9\right)}+\frac{3}{b^{2}+3b+9})$$

Cancel out b-3 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{2b+3}{-b^{2}-3b-9}+\frac{3}{b^{2}+3b+9})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of -b^{2}-3b-9 and b^{2}+3b+9 is b^{2}+3b+9. Multiply \frac{2b+3}{-b^{2}-3b-9} times \frac{-1}{-1}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{-\left(2b+3\right)}{b^{2}+3b+9}+\frac{3}{b^{2}+3b+9})$$

Since \frac{-\left(2b+3\right)}{b^{2}+3b+9} and \frac{3}{b^{2}+3b+9} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{-\left(2b+3\right)+3}{b^{2}+3b+9})$$

Do the multiplications in -\left(2b+3\right)+3.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{-2b-3+3}{b^{2}+3b+9})$$

Combine like terms in -2b-3+3.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{-2b}{b^{2}+3b+9})$$

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

$$\frac{\left(b^{2}+3b^{1}+9\right)\frac{\mathrm{d}}{\mathrm{d}b}(-2b^{1})-\left(-2b^{1}\frac{\mathrm{d}}{\mathrm{d}b}(b^{2}+3b^{1}+9)\right)}{\left(b^{2}+3b^{1}+9\right)^{2}}$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

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

Simplify.

$$\frac{\left(b^{2}+3b^{1}+9\right)\left(-2\right)b^{0}-\left(-2b^{1}\left(2b^{1}+3b^{0}\right)\right)}{\left(b^{2}+3b^{1}+9\right)^{2}}$$

Multiply b^{2}+3b^{1}+9 times -2b^{0}.

$$\frac{b^{2}\left(-2\right)b^{0}+3b^{1}\left(-2\right)b^{0}+9\left(-2\right)b^{0}-\left(-2b^{1}\left(2b^{1}+3b^{0}\right)\right)}{\left(b^{2}+3b^{1}+9\right)^{2}}$$

Multiply -2b^{1} times 2b^{1}+3b^{0}.

$$\frac{b^{2}\left(-2\right)b^{0}+3b^{1}\left(-2\right)b^{0}+9\left(-2\right)b^{0}-\left(-2b^{1}\times 2b^{1}-2b^{1}\times 3b^{0}\right)}{\left(b^{2}+3b^{1}+9\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{-2b^{2}+3\left(-2\right)b^{1}+9\left(-2\right)b^{0}-\left(-2\times 2b^{1+1}-2\times 3b^{1}\right)}{\left(b^{2}+3b^{1}+9\right)^{2}}$$

Simplify.

$$\frac{-2b^{2}-6b^{1}-18b^{0}-\left(-4b^{2}-6b^{1}\right)}{\left(b^{2}+3b^{1}+9\right)^{2}}$$

Combine like terms.

$$\frac{2b^{2}-18b^{0}}{\left(b^{2}+3b^{1}+9\right)^{2}}$$

For any term t, t^{1}=t.

$$\frac{2b^{2}-18b^{0}}{\left(b^{2}+3b+9\right)^{2}}$$

For any term t except 0, t^{0}=1.

$$\frac{2b^{2}-18}{\left(b^{2}+3b+9\right)^{2}}$$