Solve (2/3a-b)(2/3a+b)-2/3(a-b)^2+2(1/3a)^2

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

Evaluate

$$\frac{b\left(4a-5b\right)}{3}$$

Solution Steps

Consider \left(\frac{2}{3}a-b\right)\left(\frac{2}{3}a+b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(\frac{2}{3}a\right)^{2}.

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

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

$$\frac{4}{9}a^{2}-b^{2}-\frac{2}{3}\left(a-b\right)^{2}+2\times \left(\frac{1}{3}a\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$\frac{4}{9}a^{2}-b^{2}-\frac{2}{3}\left(a^{2}-2ab+b^{2}\right)+2\times \left(\frac{1}{3}a\right)^{2}$$

Use the distributive property to multiply -\frac{2}{3} by a^{2}-2ab+b^{2}.

$$\frac{4}{9}a^{2}-b^{2}-\frac{2}{3}a^{2}+\frac{4}{3}ab-\frac{2}{3}b^{2}+2\times \left(\frac{1}{3}a\right)^{2}$$

Combine \frac{4}{9}a^{2} and -\frac{2}{3}a^{2} to get -\frac{2}{9}a^{2}.

$$-\frac{2}{9}a^{2}-b^{2}+\frac{4}{3}ab-\frac{2}{3}b^{2}+2\times \left(\frac{1}{3}a\right)^{2}$$

Combine -b^{2} and -\frac{2}{3}b^{2} to get -\frac{5}{3}b^{2}.

$$-\frac{2}{9}a^{2}-\frac{5}{3}b^{2}+\frac{4}{3}ab+2\times \left(\frac{1}{3}a\right)^{2}$$

Expand \left(\frac{1}{3}a\right)^{2}.

$$-\frac{2}{9}a^{2}-\frac{5}{3}b^{2}+\frac{4}{3}ab+2\times \left(\frac{1}{3}\right)^{2}a^{2}$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$-\frac{2}{9}a^{2}-\frac{5}{3}b^{2}+\frac{4}{3}ab+2\times \frac{1}{9}a^{2}$$

Multiply 2 and \frac{1}{9} to get \frac{2}{9}.

$$-\frac{2}{9}a^{2}-\frac{5}{3}b^{2}+\frac{4}{3}ab+\frac{2}{9}a^{2}$$

Combine -\frac{2}{9}a^{2} and \frac{2}{9}a^{2} to get 0.

$$-\frac{5}{3}b^{2}+\frac{4}{3}ab$$

Expand

$$\frac{4ab-5b^{2}}{3}$$

Solution Steps

Consider \left(\frac{2}{3}a-b\right)\left(\frac{2}{3}a+b\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(\frac{2}{3}a\right)^{2}.

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

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

$$\frac{4}{9}a^{2}-b^{2}-\frac{2}{3}\left(a-b\right)^{2}+2\times \left(\frac{1}{3}a\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$\frac{4}{9}a^{2}-b^{2}-\frac{2}{3}\left(a^{2}-2ab+b^{2}\right)+2\times \left(\frac{1}{3}a\right)^{2}$$

Use the distributive property to multiply -\frac{2}{3} by a^{2}-2ab+b^{2}.

$$\frac{4}{9}a^{2}-b^{2}-\frac{2}{3}a^{2}+\frac{4}{3}ab-\frac{2}{3}b^{2}+2\times \left(\frac{1}{3}a\right)^{2}$$

Combine \frac{4}{9}a^{2} and -\frac{2}{3}a^{2} to get -\frac{2}{9}a^{2}.

$$-\frac{2}{9}a^{2}-b^{2}+\frac{4}{3}ab-\frac{2}{3}b^{2}+2\times \left(\frac{1}{3}a\right)^{2}$$

Combine -b^{2} and -\frac{2}{3}b^{2} to get -\frac{5}{3}b^{2}.

$$-\frac{2}{9}a^{2}-\frac{5}{3}b^{2}+\frac{4}{3}ab+2\times \left(\frac{1}{3}a\right)^{2}$$

Expand \left(\frac{1}{3}a\right)^{2}.

$$-\frac{2}{9}a^{2}-\frac{5}{3}b^{2}+\frac{4}{3}ab+2\times \left(\frac{1}{3}\right)^{2}a^{2}$$

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$-\frac{2}{9}a^{2}-\frac{5}{3}b^{2}+\frac{4}{3}ab+2\times \frac{1}{9}a^{2}$$

Multiply 2 and \frac{1}{9} to get \frac{2}{9}.

$$-\frac{2}{9}a^{2}-\frac{5}{3}b^{2}+\frac{4}{3}ab+\frac{2}{9}a^{2}$$

Combine -\frac{2}{9}a^{2} and \frac{2}{9}a^{2} to get 0.

$$-\frac{5}{3}b^{2}+\frac{4}{3}ab$$