Solve [(2/3)^2-(8/9+8/3)+(7/3)^2]|(frac{1}{7})^2+(frac{4}{7}-1)-(frac{2}{7})^2|

$$[ ( \frac { 2 } { 3 } ) ^ { 2 } - ( \frac { 8 } { 9 } + \frac { 8 } { 3 } ) + ( \frac { 7 } { 3 } ) ^ { 2 } ] | ( \frac { 1 } { 7 } ) ^ { 2 } + ( \frac { 4 } { 7 } - 1 ) - ( \frac { 2 } { 7 } ) ^ { 2 } |$$

Evaluate

$$\frac{8}{7}\approx 1.142857143$$

Solution Steps

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

$$\left(\frac{4}{9}-\left(\frac{8}{9}+\frac{8}{3}\right)+\left(\frac{7}{3}\right)^{2}\right)|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Least common multiple of 9 and 3 is 9. Convert \frac{8}{9} and \frac{8}{3} to fractions with denominator 9.

$$\left(\frac{4}{9}-\left(\frac{8}{9}+\frac{24}{9}\right)+\left(\frac{7}{3}\right)^{2}\right)|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Since \frac{8}{9} and \frac{24}{9} have the same denominator, add them by adding their numerators.

$$\left(\frac{4}{9}-\frac{8+24}{9}+\left(\frac{7}{3}\right)^{2}\right)|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Add 8 and 24 to get 32.

$$\left(\frac{4}{9}-\frac{32}{9}+\left(\frac{7}{3}\right)^{2}\right)|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Since \frac{4}{9} and \frac{32}{9} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{4-32}{9}+\left(\frac{7}{3}\right)^{2}\right)|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Subtract 32 from 4 to get -28.

$$\left(-\frac{28}{9}+\left(\frac{7}{3}\right)^{2}\right)|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Calculate \frac{7}{3} to the power of 2 and get \frac{49}{9}.

$$\left(-\frac{28}{9}+\frac{49}{9}\right)|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Since -\frac{28}{9} and \frac{49}{9} have the same denominator, add them by adding their numerators.

$$\frac{-28+49}{9}|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Add -28 and 49 to get 21.

$$\frac{21}{9}|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Reduce the fraction \frac{21}{9} to lowest terms by extracting and canceling out 3.

$$\frac{7}{3}|\left(\frac{1}{7}\right)^{2}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Calculate \frac{1}{7} to the power of 2 and get \frac{1}{49}.

$$\frac{7}{3}|\frac{1}{49}+\frac{4}{7}-1-\left(\frac{2}{7}\right)^{2}|$$

Least common multiple of 49 and 7 is 49. Convert \frac{1}{49} and \frac{4}{7} to fractions with denominator 49.

$$\frac{7}{3}|\frac{1}{49}+\frac{28}{49}-1-\left(\frac{2}{7}\right)^{2}|$$

Since \frac{1}{49} and \frac{28}{49} have the same denominator, add them by adding their numerators.

$$\frac{7}{3}|\frac{1+28}{49}-1-\left(\frac{2}{7}\right)^{2}|$$

Add 1 and 28 to get 29.

$$\frac{7}{3}|\frac{29}{49}-1-\left(\frac{2}{7}\right)^{2}|$$

Convert 1 to fraction \frac{49}{49}.

$$\frac{7}{3}|\frac{29}{49}-\frac{49}{49}-\left(\frac{2}{7}\right)^{2}|$$

Since \frac{29}{49} and \frac{49}{49} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{3}|\frac{29-49}{49}-\left(\frac{2}{7}\right)^{2}|$$

Subtract 49 from 29 to get -20.

$$\frac{7}{3}|-\frac{20}{49}-\left(\frac{2}{7}\right)^{2}|$$

Calculate \frac{2}{7} to the power of 2 and get \frac{4}{49}.

$$\frac{7}{3}|-\frac{20}{49}-\frac{4}{49}|$$

Since -\frac{20}{49} and \frac{4}{49} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{3}|\frac{-20-4}{49}|$$

Subtract 4 from -20 to get -24.

$$\frac{7}{3}|-\frac{24}{49}|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{24}{49} is \frac{24}{49}.

$$\frac{7}{3}\times \frac{24}{49}$$

Multiply \frac{7}{3} times \frac{24}{49} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7\times 24}{3\times 49}$$

Do the multiplications in the fraction \frac{7\times 24}{3\times 49}.

$$\frac{168}{147}$$

Reduce the fraction \frac{168}{147} to lowest terms by extracting and canceling out 21.

$$\frac{8}{7}$$

Factor

$$\frac{2 ^ {3}}{7} = 1\frac{1}{7} = 1.1428571428571428$$