Solve {[3*(18/7-2/3+1)*(2/5+1)*1/61]^3+frac{4}{25}}:frac{7}{5}

$$\{ [ 3 \cdot ( \frac { 18 } { 7 } - \frac { 2 } { 3 } + 1 ) \cdot ( \frac { 2 } { 5 } + 1 ) \cdot \frac { 1 } { 61 } ] ^ { 3 } + \frac { 4 } { 25 } \} : \frac { 7 } { 5 }$$

Evaluate

$$\frac{3}{25}=0.12$$

Solution Steps

Subtract \frac{2}{3} from \frac{18}{7} to get \frac{40}{21}.

$$\frac{\left(3\left(\frac{40}{21}+1\right)\left(\frac{2}{5}+1\right)\times \frac{1}{61}\right)^{3}+\frac{4}{25}}{\frac{7}{5}}$$

Add \frac{40}{21} and 1 to get \frac{61}{21}.

$$\frac{\left(3\times \frac{61}{21}\left(\frac{2}{5}+1\right)\times \frac{1}{61}\right)^{3}+\frac{4}{25}}{\frac{7}{5}}$$

Multiply 3 and \frac{61}{21} to get \frac{61}{7}.

$$\frac{\left(\frac{61}{7}\left(\frac{2}{5}+1\right)\times \frac{1}{61}\right)^{3}+\frac{4}{25}}{\frac{7}{5}}$$

Add \frac{2}{5} and 1 to get \frac{7}{5}.

$$\frac{\left(\frac{61}{7}\times \frac{7}{5}\times \frac{1}{61}\right)^{3}+\frac{4}{25}}{\frac{7}{5}}$$

Multiply \frac{61}{7} and \frac{7}{5} to get \frac{61}{5}.

$$\frac{\left(\frac{61}{5}\times \frac{1}{61}\right)^{3}+\frac{4}{25}}{\frac{7}{5}}$$

Multiply \frac{61}{5} and \frac{1}{61} to get \frac{1}{5}.

$$\frac{\left(\frac{1}{5}\right)^{3}+\frac{4}{25}}{\frac{7}{5}}$$

Calculate \frac{1}{5} to the power of 3 and get \frac{1}{125}.

$$\frac{\frac{1}{125}+\frac{4}{25}}{\frac{7}{5}}$$

Add \frac{1}{125} and \frac{4}{25} to get \frac{21}{125}.

$$\frac{\frac{21}{125}}{\frac{7}{5}}$$

Divide \frac{21}{125} by \frac{7}{5} by multiplying \frac{21}{125} by the reciprocal of \frac{7}{5}.

$$\frac{21}{125}\times \frac{5}{7}$$

Multiply \frac{21}{125} and \frac{5}{7} to get \frac{3}{25}.

$$\frac{3}{25}$$

Factor

$$\frac{3}{5 ^ {2}} = 0.12$$