Solve {-[(-1/4ab)^{2}*(-2b)^3+1/6ab^5]:(-1/3b^5)-2a(3a^2-5/2a+frac{1}{4})}:(frac{1}{2}a)

$$\{ - [ ( - \frac { 1 } { 4 } a b ) ^ { 2 } \cdot ( - 2 b ) ^ { 3 } + \frac { 1 } { 6 } a b ^ { 5 } ] : ( - \frac { 1 } { 3 } b ^ { 5 } ) - 2 a ( 3 a ^ { 2 } - \frac { 5 } { 2 } a + \frac { 1 } { 4 } ) \} : ( \frac { 1 } { 2 } a )$$

Evaluate

$$-a\left(12a-7\right)$$

Short Solution Steps

Expand \left(-\frac{1}{4}ab\right)^{2}.

$$\frac{\frac{-\left(\left(-\frac{1}{4}\right)^{2}a^{2}b^{2}\left(-2b\right)^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Calculate -\frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$\frac{\frac{-\left(\frac{1}{16}a^{2}b^{2}\left(-2b\right)^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Expand \left(-2b\right)^{3}.

$$\frac{\frac{-\left(\frac{1}{16}a^{2}b^{2}\left(-2\right)^{3}b^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Calculate -2 to the power of 3 and get -8.

$$\frac{\frac{-\left(\frac{1}{16}a^{2}b^{2}\left(-8\right)b^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Multiply \frac{1}{16} and -8 to get -\frac{1}{2}.

$$\frac{\frac{-\left(-\frac{1}{2}a^{2}b^{2}b^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

To multiply powers of the same base, add their exponents. Add 2 and 3 to get 5.

$$\frac{\frac{-\left(-\frac{1}{2}a^{2}b^{5}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

To find the opposite of -\frac{1}{2}a^{2}b^{5}+\frac{1}{6}ab^{5}, find the opposite of each term.

$$\frac{\frac{\frac{1}{2}a^{2}b^{5}-\frac{1}{6}ab^{5}}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Factor the expressions that are not already factored in \frac{\frac{1}{2}a^{2}b^{5}-\frac{1}{6}ab^{5}}{-\frac{1}{3}b^{5}}.

$$\frac{\frac{\frac{1}{6}a\left(3a-1\right)b^{5}}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Cancel out b^{5} in both numerator and denominator.

$$\frac{\frac{\frac{1}{6}a\left(3a-1\right)}{-\frac{1}{3}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Divide \frac{1}{6}a\left(3a-1\right) by -\frac{1}{3} to get -\frac{1}{2}a\left(3a-1\right).

$$\frac{-\frac{1}{2}a\left(3a-1\right)-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Factor the expressions that are not already factored.

$$\frac{\frac{1}{2}\left(-12a+7\right)a^{2}}{\frac{1}{2}a}$$

Cancel out a in both numerator and denominator.

$$\frac{\frac{1}{2}a\left(-12a+7\right)}{\frac{1}{2}}$$

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

$$\frac{a\left(-12a+7\right)}{\left(\frac{1}{2}\right)^{0}}$$

Expand the expression.

$$-12a^{2}+7a$$

Expand

$$7a-12a^{2}$$

Short Solution Steps

Expand \left(-\frac{1}{4}ab\right)^{2}.

$$\frac{\frac{-\left(\left(-\frac{1}{4}\right)^{2}a^{2}b^{2}\left(-2b\right)^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Calculate -\frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$\frac{\frac{-\left(\frac{1}{16}a^{2}b^{2}\left(-2b\right)^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Expand \left(-2b\right)^{3}.

$$\frac{\frac{-\left(\frac{1}{16}a^{2}b^{2}\left(-2\right)^{3}b^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Calculate -2 to the power of 3 and get -8.

$$\frac{\frac{-\left(\frac{1}{16}a^{2}b^{2}\left(-8\right)b^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Multiply \frac{1}{16} and -8 to get -\frac{1}{2}.

$$\frac{\frac{-\left(-\frac{1}{2}a^{2}b^{2}b^{3}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

To multiply powers of the same base, add their exponents. Add 2 and 3 to get 5.

$$\frac{\frac{-\left(-\frac{1}{2}a^{2}b^{5}+\frac{1}{6}ab^{5}\right)}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

To find the opposite of -\frac{1}{2}a^{2}b^{5}+\frac{1}{6}ab^{5}, find the opposite of each term.

$$\frac{\frac{\frac{1}{2}a^{2}b^{5}-\frac{1}{6}ab^{5}}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Factor the expressions that are not already factored in \frac{\frac{1}{2}a^{2}b^{5}-\frac{1}{6}ab^{5}}{-\frac{1}{3}b^{5}}.

$$\frac{\frac{\frac{1}{6}a\left(3a-1\right)b^{5}}{-\frac{1}{3}b^{5}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Cancel out b^{5} in both numerator and denominator.

$$\frac{\frac{\frac{1}{6}a\left(3a-1\right)}{-\frac{1}{3}}-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Divide \frac{1}{6}a\left(3a-1\right) by -\frac{1}{3} to get -\frac{1}{2}a\left(3a-1\right).

$$\frac{-\frac{1}{2}a\left(3a-1\right)-2a\left(3a^{2}-\frac{5}{2}a+\frac{1}{4}\right)}{\frac{1}{2}a}$$

Factor the expressions that are not already factored.

$$\frac{\frac{1}{2}\left(-12a+7\right)a^{2}}{\frac{1}{2}a}$$

Cancel out a in both numerator and denominator.

$$\frac{\frac{1}{2}a\left(-12a+7\right)}{\frac{1}{2}}$$

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

$$\frac{a\left(-12a+7\right)}{\left(\frac{1}{2}\right)^{0}}$$

Expand the expression.

$$-12a^{2}+7a$$