Solve (6a-1/3b)+(2a+2/3b)^2+(2/3b-2a)(2a+2/3b)+frac{4}{3}ab

$$( 6 a - \frac { 1 } { 3 } b ) + ( 2 a + \frac { 2 } { 3 } b ) ^ { 2 } + ( \frac { 2 } { 3 } b - 2 a ) ( 2 a + \frac { 2 } { 3 } b ) + \frac { 4 } { 3 } a b$$

Evaluate

$$4ab+\frac{8b^{2}}{9}-\frac{b}{3}+6a$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2a+\frac{2}{3}b\right)^{2}.

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\left(\frac{2}{3}b-2a\right)\left(2a+\frac{2}{3}b\right)+\frac{4}{3}ab$$

Consider \left(\frac{2}{3}b-2a\right)\left(2a+\frac{2}{3}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}.

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\left(\frac{2}{3}b\right)^{2}-\left(2a\right)^{2}+\frac{4}{3}ab$$

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

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\left(\frac{2}{3}\right)^{2}b^{2}-\left(2a\right)^{2}+\frac{4}{3}ab$$

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

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\frac{4}{9}b^{2}-\left(2a\right)^{2}+\frac{4}{3}ab$$

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

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\frac{4}{9}b^{2}-2^{2}a^{2}+\frac{4}{3}ab$$

Calculate 2 to the power of 2 and get 4.

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\frac{4}{9}b^{2}-4a^{2}+\frac{4}{3}ab$$

Combine \frac{4}{9}b^{2} and \frac{4}{9}b^{2} to get \frac{8}{9}b^{2}.

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{8}{9}b^{2}-4a^{2}+\frac{4}{3}ab$$

Combine 4a^{2} and -4a^{2} to get 0.

$$6a-\frac{1}{3}b+\frac{8}{3}ab+\frac{8}{9}b^{2}+\frac{4}{3}ab$$

Combine \frac{8}{3}ab and \frac{4}{3}ab to get 4ab.

$$6a-\frac{1}{3}b+4ab+\frac{8}{9}b^{2}$$

Expand

$$4ab+\frac{8b^{2}}{9}-\frac{b}{3}+6a$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(2a+\frac{2}{3}b\right)^{2}.

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\left(\frac{2}{3}b-2a\right)\left(2a+\frac{2}{3}b\right)+\frac{4}{3}ab$$

Consider \left(\frac{2}{3}b-2a\right)\left(2a+\frac{2}{3}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}.

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\left(\frac{2}{3}b\right)^{2}-\left(2a\right)^{2}+\frac{4}{3}ab$$

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

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\left(\frac{2}{3}\right)^{2}b^{2}-\left(2a\right)^{2}+\frac{4}{3}ab$$

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

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\frac{4}{9}b^{2}-\left(2a\right)^{2}+\frac{4}{3}ab$$

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

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\frac{4}{9}b^{2}-2^{2}a^{2}+\frac{4}{3}ab$$

Calculate 2 to the power of 2 and get 4.

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{4}{9}b^{2}+\frac{4}{9}b^{2}-4a^{2}+\frac{4}{3}ab$$

Combine \frac{4}{9}b^{2} and \frac{4}{9}b^{2} to get \frac{8}{9}b^{2}.

$$6a-\frac{1}{3}b+4a^{2}+\frac{8}{3}ab+\frac{8}{9}b^{2}-4a^{2}+\frac{4}{3}ab$$

Combine 4a^{2} and -4a^{2} to get 0.

$$6a-\frac{1}{3}b+\frac{8}{3}ab+\frac{8}{9}b^{2}+\frac{4}{3}ab$$

Combine \frac{8}{3}ab and \frac{4}{3}ab to get 4ab.

$$6a-\frac{1}{3}b+4ab+\frac{8}{9}b^{2}$$