Solve (6^-1+(6a)^-1-2a^-2):(a-3/18a^-2+2a)

$$( 6 ^ { - 1 } + ( 6 a ) ^ { - 1 } - 2 a ^ { - 2 } ) : ( \frac { a - 3 } { 18 a ^ { - 2 } + 2 a } )$$

Evaluate

$$\frac{\left(a+4\right)\left(a^{3}+9\right)}{3a^{4}}$$

Short Solution Steps

Calculate 6 to the power of -1 and get \frac{1}{6}.

$$\frac{\frac{1}{6}+\left(6a\right)^{-1}-2a^{-2}}{\frac{a-3}{18a^{-2}+2a}}$$

Expand \left(6a\right)^{-1}.

$$\frac{\frac{1}{6}+6^{-1}a^{-1}-2a^{-2}}{\frac{a-3}{18a^{-2}+2a}}$$

Calculate 6 to the power of -1 and get \frac{1}{6}.

$$\frac{\frac{1}{6}+\frac{1}{6}a^{-1}-2a^{-2}}{\frac{a-3}{18a^{-2}+2a}}$$

Divide \frac{1}{6}+\frac{1}{6}a^{-1}-2a^{-2} by \frac{a-3}{18a^{-2}+2a} by multiplying \frac{1}{6}+\frac{1}{6}a^{-1}-2a^{-2} by the reciprocal of \frac{a-3}{18a^{-2}+2a}.

$$\frac{\left(\frac{1}{6}+\frac{1}{6}a^{-1}-2a^{-2}\right)\left(18a^{-2}+2a\right)}{a-3}$$

Factor the expressions that are not already factored.

$$\frac{\frac{1}{6}\times 2\left(a^{-2}\right)^{2}\left(a-3\right)\left(a+4\right)\left(a^{3}+9\right)}{a-3}$$

Cancel out a-3 in both numerator and denominator.

$$\frac{1}{6}\times 2\left(a^{-2}\right)^{2}\left(a+4\right)\left(a^{3}+9\right)$$

Expand the expression.

$$\frac{1}{3}+\frac{4}{3}\times \frac{1}{a}+3a^{-3}+12a^{-4}$$

Multiply \frac{4}{3} times \frac{1}{a} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{3}+\frac{4}{3a}+3a^{-3}+12a^{-4}$$

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

$$\frac{a}{3a}+\frac{4}{3a}+3a^{-3}+12a^{-4}$$

Since \frac{a}{3a} and \frac{4}{3a} have the same denominator, add them by adding their numerators.

$$\frac{a+4}{3a}+3a^{-3}+12a^{-4}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3a^{-3}+12a^{-4} times \frac{3a}{3a}.

$$\frac{a+4}{3a}+\frac{\left(3a^{-3}+12a^{-4}\right)\times 3a}{3a}$$

Since \frac{a+4}{3a} and \frac{\left(3a^{-3}+12a^{-4}\right)\times 3a}{3a} have the same denominator, add them by adding their numerators.

$$\frac{a+4+\left(3a^{-3}+12a^{-4}\right)\times 3a}{3a}$$

Do the multiplications in a+4+\left(3a^{-3}+12a^{-4}\right)\times 3a.

$$\frac{a+4+9a^{-2}+36a^{-3}}{3a}$$

Combine like terms in a+4+9a^{-2}+36a^{-3}.

$$\frac{a+4+\left(36+9a\right)a^{-3}}{3a}$$

Factor the expressions that are not already factored in \frac{a+4+\left(36+9a\right)a^{-3}}{3a}.

$$\frac{a^{-3}\left(a+4\right)\left(a^{3}+9\right)}{3a}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{\left(a+4\right)\left(a^{3}+9\right)}{3a^{4}}$$

Use the distributive property to multiply a+4 by a^{3}+9.

$$\frac{a^{4}+9a+4a^{3}+36}{3a^{4}}$$

Expand

$$\frac{1}{3}+\frac{4}{3a}+\frac{3}{a^{3}}+\frac{12}{a^{4}}$$

Short Solution Steps

Calculate 6 to the power of -1 and get \frac{1}{6}.

$$\frac{\frac{1}{6}+\left(6a\right)^{-1}-2a^{-2}}{\frac{a-3}{18a^{-2}+2a}}$$

Expand \left(6a\right)^{-1}.

$$\frac{\frac{1}{6}+6^{-1}a^{-1}-2a^{-2}}{\frac{a-3}{18a^{-2}+2a}}$$

Calculate 6 to the power of -1 and get \frac{1}{6}.

$$\frac{\frac{1}{6}+\frac{1}{6}a^{-1}-2a^{-2}}{\frac{a-3}{18a^{-2}+2a}}$$

Divide \frac{1}{6}+\frac{1}{6}a^{-1}-2a^{-2} by \frac{a-3}{18a^{-2}+2a} by multiplying \frac{1}{6}+\frac{1}{6}a^{-1}-2a^{-2} by the reciprocal of \frac{a-3}{18a^{-2}+2a}.

$$\frac{\left(\frac{1}{6}+\frac{1}{6}a^{-1}-2a^{-2}\right)\left(18a^{-2}+2a\right)}{a-3}$$

Factor the expressions that are not already factored.

$$\frac{\frac{1}{6}\times 2\left(a^{-2}\right)^{2}\left(a-3\right)\left(a+4\right)\left(a^{3}+9\right)}{a-3}$$

Cancel out a-3 in both numerator and denominator.

$$\frac{1}{6}\times 2\left(a^{-2}\right)^{2}\left(a+4\right)\left(a^{3}+9\right)$$

Expand the expression.

$$\frac{1}{3}+\frac{4}{3}\times \frac{1}{a}+3a^{-3}+12a^{-4}$$

Multiply \frac{4}{3} times \frac{1}{a} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{3}+\frac{4}{3a}+3a^{-3}+12a^{-4}$$

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

$$\frac{a}{3a}+\frac{4}{3a}+3a^{-3}+12a^{-4}$$

Since \frac{a}{3a} and \frac{4}{3a} have the same denominator, add them by adding their numerators.

$$\frac{a+4}{3a}+3a^{-3}+12a^{-4}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3a^{-3}+12a^{-4} times \frac{3a}{3a}.

$$\frac{a+4}{3a}+\frac{\left(3a^{-3}+12a^{-4}\right)\times 3a}{3a}$$

Since \frac{a+4}{3a} and \frac{\left(3a^{-3}+12a^{-4}\right)\times 3a}{3a} have the same denominator, add them by adding their numerators.

$$\frac{a+4+\left(3a^{-3}+12a^{-4}\right)\times 3a}{3a}$$

Do the multiplications in a+4+\left(3a^{-3}+12a^{-4}\right)\times 3a.

$$\frac{a+4+9a^{-2}+36a^{-3}}{3a}$$

Combine like terms in a+4+9a^{-2}+36a^{-3}.

$$\frac{a+4+\left(36+9a\right)a^{-3}}{3a}$$

Factor the expressions that are not already factored in \frac{a+4+\left(36+9a\right)a^{-3}}{3a}.

$$\frac{a^{-3}\left(a+4\right)\left(a^{3}+9\right)}{3a}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{\left(a+4\right)\left(a^{3}+9\right)}{3a^{4}}$$

Use the distributive property to multiply a+4 by a^{3}+9.

$$\frac{a^{4}+9a+4a^{3}+36}{3a^{4}}$$