Solve {[(2/3)^3+(1/2)^2*2/3+frac{32/135]*frac{25}{3}-4}*frac{77}{3}}{[(frac{2}{3})^2+2*frac{2}{3}*frac{1}{4}+(frac{1}{4})^2]*3}-13

$$\frac { \{ [ ( \frac { 2 } { 3 } ) ^ { 3 } + ( \frac { 1 } { 2 } ) ^ { 2 } \cdot \frac { 2 } { 3 } + \frac { 32 } { 135 } ] \cdot \frac { 25 } { 3 } - 4 \} \cdot \frac { 77 } { 3 } } { [ ( \frac { 2 } { 3 } ) ^ { 2 } + 2 \cdot \frac { 2 } { 3 } \cdot \frac { 1 } { 4 } + ( \frac { 1 } { 4 } ) ^ { 2 } ] \cdot 3 } - 13$$

Evaluate

$$\frac{17}{3}\approx 5.666666667$$

Solution Steps

Calculate \frac{2}{3} to the power of 3 and get \frac{8}{27}.

$$\frac{\left(\left(\frac{8}{27}+\left(\frac{1}{2}\right)^{2}\times \frac{2}{3}+\frac{32}{135}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

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

$$\frac{\left(\left(\frac{8}{27}+\frac{1}{4}\times \frac{2}{3}+\frac{32}{135}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Multiply \frac{1}{4} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(\left(\frac{8}{27}+\frac{1\times 2}{4\times 3}+\frac{32}{135}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Do the multiplications in the fraction \frac{1\times 2}{4\times 3}.

$$\frac{\left(\left(\frac{8}{27}+\frac{2}{12}+\frac{32}{135}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.

$$\frac{\left(\left(\frac{8}{27}+\frac{1}{6}+\frac{32}{135}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Least common multiple of 27 and 6 is 54. Convert \frac{8}{27} and \frac{1}{6} to fractions with denominator 54.

$$\frac{\left(\left(\frac{16}{54}+\frac{9}{54}+\frac{32}{135}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Since \frac{16}{54} and \frac{9}{54} have the same denominator, add them by adding their numerators.

$$\frac{\left(\left(\frac{16+9}{54}+\frac{32}{135}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Add 16 and 9 to get 25.

$$\frac{\left(\left(\frac{25}{54}+\frac{32}{135}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Least common multiple of 54 and 135 is 270. Convert \frac{25}{54} and \frac{32}{135} to fractions with denominator 270.

$$\frac{\left(\left(\frac{125}{270}+\frac{64}{270}\right)\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Since \frac{125}{270} and \frac{64}{270} have the same denominator, add them by adding their numerators.

$$\frac{\left(\frac{125+64}{270}\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Add 125 and 64 to get 189.

$$\frac{\left(\frac{189}{270}\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Reduce the fraction \frac{189}{270} to lowest terms by extracting and canceling out 27.

$$\frac{\left(\frac{7}{10}\times \frac{25}{3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Multiply \frac{7}{10} times \frac{25}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(\frac{7\times 25}{10\times 3}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Do the multiplications in the fraction \frac{7\times 25}{10\times 3}.

$$\frac{\left(\frac{175}{30}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Reduce the fraction \frac{175}{30} to lowest terms by extracting and canceling out 5.

$$\frac{\left(\frac{35}{6}-4\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Convert 4 to fraction \frac{24}{6}.

$$\frac{\left(\frac{35}{6}-\frac{24}{6}\right)\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Since \frac{35}{6} and \frac{24}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{35-24}{6}\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Subtract 24 from 35 to get 11.

$$\frac{\frac{11}{6}\times \frac{77}{3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Multiply \frac{11}{6} times \frac{77}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{11\times 77}{6\times 3}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Do the multiplications in the fraction \frac{11\times 77}{6\times 3}.

$$\frac{\frac{847}{18}}{\left(\left(\frac{2}{3}\right)^{2}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Calculate \frac{2}{3} to the power of 2 and get \frac{4}{9}.

$$\frac{\frac{847}{18}}{\left(\frac{4}{9}+2\times \frac{2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Express 2\times \frac{2}{3} as a single fraction.

$$\frac{\frac{847}{18}}{\left(\frac{4}{9}+\frac{2\times 2}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Multiply 2 and 2 to get 4.

$$\frac{\frac{847}{18}}{\left(\frac{4}{9}+\frac{4}{3}\times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

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

$$\frac{\frac{847}{18}}{\left(\frac{4}{9}+\frac{4\times 1}{3\times 4}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Cancel out 4 in both numerator and denominator.

$$\frac{\frac{847}{18}}{\left(\frac{4}{9}+\frac{1}{3}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Least common multiple of 9 and 3 is 9. Convert \frac{4}{9} and \frac{1}{3} to fractions with denominator 9.

$$\frac{\frac{847}{18}}{\left(\frac{4}{9}+\frac{3}{9}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Since \frac{4}{9} and \frac{3}{9} have the same denominator, add them by adding their numerators.

$$\frac{\frac{847}{18}}{\left(\frac{4+3}{9}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

Add 4 and 3 to get 7.

$$\frac{\frac{847}{18}}{\left(\frac{7}{9}+\left(\frac{1}{4}\right)^{2}\right)\times 3}-13$$

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

$$\frac{\frac{847}{18}}{\left(\frac{7}{9}+\frac{1}{16}\right)\times 3}-13$$

Least common multiple of 9 and 16 is 144. Convert \frac{7}{9} and \frac{1}{16} to fractions with denominator 144.

$$\frac{\frac{847}{18}}{\left(\frac{112}{144}+\frac{9}{144}\right)\times 3}-13$$

Since \frac{112}{144} and \frac{9}{144} have the same denominator, add them by adding their numerators.

$$\frac{\frac{847}{18}}{\frac{112+9}{144}\times 3}-13$$

Add 112 and 9 to get 121.

$$\frac{\frac{847}{18}}{\frac{121}{144}\times 3}-13$$

Express \frac{121}{144}\times 3 as a single fraction.

$$\frac{\frac{847}{18}}{\frac{121\times 3}{144}}-13$$

Multiply 121 and 3 to get 363.

$$\frac{\frac{847}{18}}{\frac{363}{144}}-13$$

Reduce the fraction \frac{363}{144} to lowest terms by extracting and canceling out 3.

$$\frac{\frac{847}{18}}{\frac{121}{48}}-13$$

Divide \frac{847}{18} by \frac{121}{48} by multiplying \frac{847}{18} by the reciprocal of \frac{121}{48}.

$$\frac{847}{18}\times \frac{48}{121}-13$$

Multiply \frac{847}{18} times \frac{48}{121} by multiplying numerator times numerator and denominator times denominator.

$$\frac{847\times 48}{18\times 121}-13$$

Do the multiplications in the fraction \frac{847\times 48}{18\times 121}.

$$\frac{40656}{2178}-13$$

Reduce the fraction \frac{40656}{2178} to lowest terms by extracting and canceling out 726.

$$\frac{56}{3}-13$$

Convert 13 to fraction \frac{39}{3}.

$$\frac{56}{3}-\frac{39}{3}$$

Since \frac{56}{3} and \frac{39}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{56-39}{3}$$

Subtract 39 from 56 to get 17.

$$\frac{17}{3}$$

Factor

$$\frac{17}{3} = 5\frac{2}{3} = 5.666666666666667$$