Solve [(1/3a^{2}b+2/5a^{2}b):(+11/5a^2)+2/3b]^2-(frac{1}{3}a^2+b^2)

$$[ ( \frac { 1 } { 3 } a ^ { 2 } b + \frac { 2 } { 5 } a ^ { 2 } b ) : ( + \frac { 11 } { 5 } a ^ { 2 } ) + \frac { 2 } { 3 } b ] ^ { 2 } - ( \frac { 1 } { 3 } a ^ { 2 } + b ^ { 2 } )$$

Evaluate

$$-\frac{a^{2}}{3}$$

Solution Steps

Combine \frac{1}{3}a^{2}b and \frac{2}{5}a^{2}b to get \frac{11}{15}a^{2}b.

$$\left(\frac{\frac{11}{15}a^{2}b}{\frac{11}{5}a^{2}}+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Cancel out a^{2} in both numerator and denominator.

$$\left(\frac{\frac{11}{15}b}{\frac{11}{5}}+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Divide \frac{11}{15}b by \frac{11}{5} by multiplying \frac{11}{15}b by the reciprocal of \frac{11}{5}.

$$\left(\frac{\frac{11}{15}b\times 5}{11}+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Multiply \frac{11}{15} and 5 to get \frac{11}{3}.

$$\left(\frac{\frac{11}{3}b}{11}+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Divide \frac{11}{3}b by 11 to get \frac{1}{3}b.

$$\left(\frac{1}{3}b+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Combine \frac{1}{3}b and \frac{2}{3}b to get b.

$$b^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

To find the opposite of \frac{1}{3}a^{2}+b^{2}, find the opposite of each term.

$$b^{2}-\frac{1}{3}a^{2}-b^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$-\frac{1}{3}a^{2}$$

Expand

$$-\frac{a^{2}}{3}$$

Solution Steps

Combine \frac{1}{3}a^{2}b and \frac{2}{5}a^{2}b to get \frac{11}{15}a^{2}b.

$$\left(\frac{\frac{11}{15}a^{2}b}{\frac{11}{5}a^{2}}+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Cancel out a^{2} in both numerator and denominator.

$$\left(\frac{\frac{11}{15}b}{\frac{11}{5}}+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Divide \frac{11}{15}b by \frac{11}{5} by multiplying \frac{11}{15}b by the reciprocal of \frac{11}{5}.

$$\left(\frac{\frac{11}{15}b\times 5}{11}+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Multiply \frac{11}{15} and 5 to get \frac{11}{3}.

$$\left(\frac{\frac{11}{3}b}{11}+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Divide \frac{11}{3}b by 11 to get \frac{1}{3}b.

$$\left(\frac{1}{3}b+\frac{2}{3}b\right)^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

Combine \frac{1}{3}b and \frac{2}{3}b to get b.

$$b^{2}-\left(\frac{1}{3}a^{2}+b^{2}\right)$$

To find the opposite of \frac{1}{3}a^{2}+b^{2}, find the opposite of each term.

$$b^{2}-\frac{1}{3}a^{2}-b^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$-\frac{1}{3}a^{2}$$