Solve 7/6b^{3}y^{3}-(12/7b^2y)*(-35/36by^2)+3/4by[-frac{3}{4}by(frac{3}{4}by-frac{7}{6}by)-frac{1}{16}b^2y^2]

$$\frac { 7 } { 6 } b ^ { 3 } y ^ { 3 } - ( \frac { 12 } { 7 } b ^ { 2 } y ) \cdot ( - \frac { 35 } { 36 } b y ^ { 2 } ) + \frac { 3 } { 4 } b y [ - \frac { 3 } { 4 } b y ( \frac { 3 } { 4 } b y - \frac { 7 } { 6 } b y ) - \frac { 1 } { 16 } b ^ { 2 } y ^ { 2 } ]$$

Evaluate

$$\frac{145\left(by\right)^{3}}{48}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{7}{6}b^{3}y^{3}-\frac{12}{7}b^{3}y\left(-\frac{35}{36}\right)y^{2}+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{7}{6}b^{3}y^{3}-\frac{12}{7}b^{3}y^{3}\left(-\frac{35}{36}\right)+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

Multiply \frac{12}{7} and -\frac{35}{36} to get -\frac{5}{3}.

$$\frac{7}{6}b^{3}y^{3}-\left(-\frac{5}{3}b^{3}y^{3}\right)+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

The opposite of -\frac{5}{3}b^{3}y^{3} is \frac{5}{3}b^{3}y^{3}.

$$\frac{7}{6}b^{3}y^{3}+\frac{5}{3}b^{3}y^{3}+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

Combine \frac{7}{6}b^{3}y^{3} and \frac{5}{3}b^{3}y^{3} to get \frac{17}{6}b^{3}y^{3}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

Combine \frac{3}{4}by and -\frac{7}{6}by to get -\frac{5}{12}by.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(-\frac{3}{4}by\left(-\frac{5}{12}\right)by-\frac{1}{16}b^{2}y^{2}\right)$$

Multiply -\frac{3}{4} and -\frac{5}{12} to get \frac{5}{16}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(\frac{5}{16}byby-\frac{1}{16}b^{2}y^{2}\right)$$

Multiply b and b to get b^{2}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(\frac{5}{16}b^{2}yy-\frac{1}{16}b^{2}y^{2}\right)$$

Multiply y and y to get y^{2}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(\frac{5}{16}b^{2}y^{2}-\frac{1}{16}b^{2}y^{2}\right)$$

Combine \frac{5}{16}b^{2}y^{2} and -\frac{1}{16}b^{2}y^{2} to get \frac{1}{4}b^{2}y^{2}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\times \frac{1}{4}b^{2}y^{2}$$

Multiply \frac{3}{4} and \frac{1}{4} to get \frac{3}{16}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{16}byb^{2}y^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{16}b^{3}yy^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{16}b^{3}y^{3}$$

Combine \frac{17}{6}b^{3}y^{3} and \frac{3}{16}b^{3}y^{3} to get \frac{145}{48}b^{3}y^{3}.

$$\frac{145}{48}b^{3}y^{3}$$

Expand

$$\frac{145\left(by\right)^{3}}{48}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{7}{6}b^{3}y^{3}-\frac{12}{7}b^{3}y\left(-\frac{35}{36}\right)y^{2}+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{7}{6}b^{3}y^{3}-\frac{12}{7}b^{3}y^{3}\left(-\frac{35}{36}\right)+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

Multiply \frac{12}{7} and -\frac{35}{36} to get -\frac{5}{3}.

$$\frac{7}{6}b^{3}y^{3}-\left(-\frac{5}{3}b^{3}y^{3}\right)+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

The opposite of -\frac{5}{3}b^{3}y^{3} is \frac{5}{3}b^{3}y^{3}.

$$\frac{7}{6}b^{3}y^{3}+\frac{5}{3}b^{3}y^{3}+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

Combine \frac{7}{6}b^{3}y^{3} and \frac{5}{3}b^{3}y^{3} to get \frac{17}{6}b^{3}y^{3}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(-\frac{3}{4}by\left(\frac{3}{4}by-\frac{7}{6}by\right)-\frac{1}{16}b^{2}y^{2}\right)$$

Combine \frac{3}{4}by and -\frac{7}{6}by to get -\frac{5}{12}by.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(-\frac{3}{4}by\left(-\frac{5}{12}\right)by-\frac{1}{16}b^{2}y^{2}\right)$$

Multiply -\frac{3}{4} and -\frac{5}{12} to get \frac{5}{16}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(\frac{5}{16}byby-\frac{1}{16}b^{2}y^{2}\right)$$

Multiply b and b to get b^{2}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(\frac{5}{16}b^{2}yy-\frac{1}{16}b^{2}y^{2}\right)$$

Multiply y and y to get y^{2}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\left(\frac{5}{16}b^{2}y^{2}-\frac{1}{16}b^{2}y^{2}\right)$$

Combine \frac{5}{16}b^{2}y^{2} and -\frac{1}{16}b^{2}y^{2} to get \frac{1}{4}b^{2}y^{2}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{4}by\times \frac{1}{4}b^{2}y^{2}$$

Multiply \frac{3}{4} and \frac{1}{4} to get \frac{3}{16}.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{16}byb^{2}y^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{16}b^{3}yy^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{17}{6}b^{3}y^{3}+\frac{3}{16}b^{3}y^{3}$$

Combine \frac{17}{6}b^{3}y^{3} and \frac{3}{16}b^{3}y^{3} to get \frac{145}{48}b^{3}y^{3}.

$$\frac{145}{48}b^{3}y^{3}$$