Solve (-2)^3*{[(7/8+4/15-1):(1/6-1-1/12+frac{1}{3}-frac{5}{6})+(-frac{1}{2})^3:(-frac{1}{2})^2]+(frac{3}{5}-frac{5}{8}+frac{3}{10})}

$$( - 2 ) ^ { 3 } \cdot \{ [ ( \frac { 7 } { 8 } + \frac { 4 } { 15 } - 1 ) : ( \frac { 1 } { 6 } - 1 - \frac { 1 } { 12 } + \frac { 1 } { 3 } - \frac { 5 } { 6 } ) + ( - \frac { 1 } { 2 } ) ^ { 3 } : ( - \frac { 1 } { 2 } ) ^ { 2 } ] + ( \frac { 3 } { 5 } - \frac { 5 } { 8 } + \frac { 3 } { 10 } ) \}$$

Evaluate

$$\frac{13}{5}=2.6$$

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 2 from 3 to get 1.

$$\left(-2\right)^{3}\left(\frac{\frac{7}{8}+\frac{4}{15}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

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

$$-8\left(\frac{\frac{7}{8}+\frac{4}{15}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 8 and 15 is 120. Convert \frac{7}{8} and \frac{4}{15} to fractions with denominator 120.

$$-8\left(\frac{\frac{105}{120}+\frac{32}{120}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since \frac{105}{120} and \frac{32}{120} have the same denominator, add them by adding their numerators.

$$-8\left(\frac{\frac{105+32}{120}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Add 105 and 32 to get 137.

$$-8\left(\frac{\frac{137}{120}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Convert 1 to fraction \frac{120}{120}.

$$-8\left(\frac{\frac{137}{120}-\frac{120}{120}}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since \frac{137}{120} and \frac{120}{120} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{\frac{137-120}{120}}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 120 from 137 to get 17.

$$-8\left(\frac{\frac{17}{120}}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Convert 1 to fraction \frac{6}{6}.

$$-8\left(\frac{\frac{17}{120}}{\frac{1}{6}-\frac{6}{6}-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since \frac{1}{6} and \frac{6}{6} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{\frac{17}{120}}{\frac{1-6}{6}-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 6 from 1 to get -5.

$$-8\left(\frac{\frac{17}{120}}{-\frac{5}{6}-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 6 and 12 is 12. Convert -\frac{5}{6} and \frac{1}{12} to fractions with denominator 12.

$$-8\left(\frac{\frac{17}{120}}{-\frac{10}{12}-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since -\frac{10}{12} and \frac{1}{12} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{\frac{17}{120}}{\frac{-10-1}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 1 from -10 to get -11.

$$-8\left(\frac{\frac{17}{120}}{-\frac{11}{12}+\frac{1}{3}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 12 and 3 is 12. Convert -\frac{11}{12} and \frac{1}{3} to fractions with denominator 12.

$$-8\left(\frac{\frac{17}{120}}{-\frac{11}{12}+\frac{4}{12}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since -\frac{11}{12} and \frac{4}{12} have the same denominator, add them by adding their numerators.

$$-8\left(\frac{\frac{17}{120}}{\frac{-11+4}{12}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Add -11 and 4 to get -7.

$$-8\left(\frac{\frac{17}{120}}{-\frac{7}{12}-\frac{5}{6}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 12 and 6 is 12. Convert -\frac{7}{12} and \frac{5}{6} to fractions with denominator 12.

$$-8\left(\frac{\frac{17}{120}}{-\frac{7}{12}-\frac{10}{12}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since -\frac{7}{12} and \frac{10}{12} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{\frac{17}{120}}{\frac{-7-10}{12}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 10 from -7 to get -17.

$$-8\left(\frac{\frac{17}{120}}{-\frac{17}{12}}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Divide \frac{17}{120} by -\frac{17}{12} by multiplying \frac{17}{120} by the reciprocal of -\frac{17}{12}.

$$-8\left(\frac{17}{120}\left(-\frac{12}{17}\right)+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Multiply \frac{17}{120} times -\frac{12}{17} by multiplying numerator times numerator and denominator times denominator.

$$-8\left(\frac{17\left(-12\right)}{120\times 17}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Cancel out 17 in both numerator and denominator.

$$-8\left(\frac{-12}{120}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Reduce the fraction \frac{-12}{120} to lowest terms by extracting and canceling out 12.

$$-8\left(-\frac{1}{10}+\left(-\frac{1}{2}\right)^{1}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

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

$$-8\left(-\frac{1}{10}-\frac{1}{2}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 10 and 2 is 10. Convert -\frac{1}{10} and \frac{1}{2} to fractions with denominator 10.

$$-8\left(-\frac{1}{10}-\frac{5}{10}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since -\frac{1}{10} and \frac{5}{10} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{-1-5}{10}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 5 from -1 to get -6.

$$-8\left(\frac{-6}{10}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

$$-8\left(-\frac{3}{5}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Add -\frac{3}{5} and \frac{3}{5} to get 0.

$$-8\left(-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 8 and 10 is 40. Convert -\frac{5}{8} and \frac{3}{10} to fractions with denominator 40.

$$-8\left(-\frac{25}{40}+\frac{12}{40}\right)$$

Since -\frac{25}{40} and \frac{12}{40} have the same denominator, add them by adding their numerators.

$$-8\times \frac{-25+12}{40}$$

Add -25 and 12 to get -13.

$$-8\left(-\frac{13}{40}\right)$$

Express -8\left(-\frac{13}{40}\right) as a single fraction.

$$\frac{-8\left(-13\right)}{40}$$

Multiply -8 and -13 to get 104.

$$\frac{104}{40}$$

Reduce the fraction \frac{104}{40} to lowest terms by extracting and canceling out 8.

$$\frac{13}{5}$$

Factor

$$\frac{13}{5} = 2\frac{3}{5} = 2.6$$