Solve (y/2+7/3)^3-(-1/2-7/3)^3

$$( \frac { y } { 2 } + \frac { 7 } { 3 } ) ^ { 3 } - ( - \frac { 1 } { 2 } - \frac { 7 } { 3 } ) ^ { 3 }$$

Evaluate

$$\frac{y^{3}}{8}+\frac{7y^{2}}{4}+\frac{49y}{6}+\frac{7657}{216}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 3 is 6. Multiply \frac{y}{2} times \frac{3}{3}. Multiply \frac{7}{3} times \frac{2}{2}.

$$\left(\frac{3y}{6}+\frac{7\times 2}{6}\right)^{3}-\left(-\frac{1}{2}-\frac{7}{3}\right)^{3}$$

Since \frac{3y}{6} and \frac{7\times 2}{6} have the same denominator, add them by adding their numerators.

$$\left(\frac{3y+7\times 2}{6}\right)^{3}-\left(-\frac{1}{2}-\frac{7}{3}\right)^{3}$$

Do the multiplications in 3y+7\times 2.

$$\left(\frac{3y+14}{6}\right)^{3}-\left(-\frac{1}{2}-\frac{7}{3}\right)^{3}$$

To raise \frac{3y+14}{6} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3y+14\right)^{3}}{6^{3}}-\left(-\frac{1}{2}-\frac{7}{3}\right)^{3}$$

Subtract \frac{7}{3} from -\frac{1}{2} to get -\frac{17}{6}.

$$\frac{\left(3y+14\right)^{3}}{6^{3}}-\left(-\frac{17}{6}\right)^{3}$$

Calculate -\frac{17}{6} to the power of 3 and get -\frac{4913}{216}.

$$\frac{\left(3y+14\right)^{3}}{6^{3}}-\left(-\frac{4913}{216}\right)$$

The opposite of -\frac{4913}{216} is \frac{4913}{216}.

$$\frac{\left(3y+14\right)^{3}}{6^{3}}+\frac{4913}{216}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 6^{3}.

$$\frac{\left(3y+14\right)^{3}}{216}+\frac{4913}{216}$$

Since \frac{\left(3y+14\right)^{3}}{216} and \frac{4913}{216} have the same denominator, add them by adding their numerators.

$$\frac{\left(3y+14\right)^{3}+4913}{216}$$

Do the multiplications in \left(3y+14\right)^{3}+4913.

$$\frac{\left(3y\right)^{3}+3\times \left(3y\right)^{2}\times 14+3\times 3y\times 14^{2}+14^{3}+4913}{216}$$

Combine like terms in \left(3y\right)^{3}+3\times \left(3y\right)^{2}\times 14+3\times 3y\times 14^{2}+14^{3}+4913.

$$\frac{27y^{3}+378y^{2}+1764y+7657}{216}$$

Expand

$$\frac{y^{3}}{8}+\frac{7y^{2}}{4}+\frac{49y}{6}+\frac{7657}{216}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 3 is 6. Multiply \frac{y}{2} times \frac{3}{3}. Multiply \frac{7}{3} times \frac{2}{2}.

$$\left(\frac{3y}{6}+\frac{7\times 2}{6}\right)^{3}-\left(-\frac{1}{2}-\frac{7}{3}\right)^{3}$$

Since \frac{3y}{6} and \frac{7\times 2}{6} have the same denominator, add them by adding their numerators.

$$\left(\frac{3y+7\times 2}{6}\right)^{3}-\left(-\frac{1}{2}-\frac{7}{3}\right)^{3}$$

Do the multiplications in 3y+7\times 2.

$$\left(\frac{3y+14}{6}\right)^{3}-\left(-\frac{1}{2}-\frac{7}{3}\right)^{3}$$

To raise \frac{3y+14}{6} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3y+14\right)^{3}}{6^{3}}-\left(-\frac{1}{2}-\frac{7}{3}\right)^{3}$$

Subtract \frac{7}{3} from -\frac{1}{2} to get -\frac{17}{6}.

$$\frac{\left(3y+14\right)^{3}}{6^{3}}-\left(-\frac{17}{6}\right)^{3}$$

Calculate -\frac{17}{6} to the power of 3 and get -\frac{4913}{216}.

$$\frac{\left(3y+14\right)^{3}}{6^{3}}-\left(-\frac{4913}{216}\right)$$

The opposite of -\frac{4913}{216} is \frac{4913}{216}.

$$\frac{\left(3y+14\right)^{3}}{6^{3}}+\frac{4913}{216}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 6^{3}.

$$\frac{\left(3y+14\right)^{3}}{216}+\frac{4913}{216}$$

Since \frac{\left(3y+14\right)^{3}}{216} and \frac{4913}{216} have the same denominator, add them by adding their numerators.

$$\frac{\left(3y+14\right)^{3}+4913}{216}$$

Do the multiplications in \left(3y+14\right)^{3}+4913.

$$\frac{\left(3y\right)^{3}+3\times \left(3y\right)^{2}\times 14+3\times 3y\times 14^{2}+14^{3}+4913}{216}$$

Combine like terms in \left(3y\right)^{3}+3\times \left(3y\right)^{2}\times 14+3\times 3y\times 14^{2}+14^{3}+4913.

$$\frac{27y^{3}+378y^{2}+1764y+7657}{216}$$