Solve {l}{[(2/3x+1/2y)^2-(2/3x+1/2y)(frac{4}{9}x^2+frac{1}{4}y^2-frac{1}{3}xy)]+2x+(frac{1}{3}x+frac{1}{2}y)(frac{1}{3}x-frac{1}{2}y)+}{+frac{2}{3}x(y-frac{1}{6}x)}

$$\left. \begin{array} { l } { [ ( \frac { 2 } { 3 } x + \frac { 1 } { 2 } y ) ^ { 2 } - ( \frac { 2 } { 3 } x + \frac { 1 } { 2 } y ) ( \frac { 4 } { 9 } x ^ { 2 } + \frac { 1 } { 4 } y ^ { 2 } - \frac { 1 } { 3 } x y ) ] + 2 x + ( \frac { 1 } { 3 } x + \frac { 1 } { 2 } y ) ( \frac { 1 } { 3 } x - \frac { 1 } { 2 } y ) + } \\ { + \frac { 2 } { 3 } x ( y - \frac { 1 } { 6 } x ) } \end{array} \right.$$

Evaluate

$$-\frac{8x^{3}}{27}-\frac{y^{3}}{8}+\frac{4xy}{3}+\frac{4x^{2}}{9}+2x$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{2}{3}x+\frac{1}{2}y\right)^{2}.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\left(\frac{2}{3}x+\frac{1}{2}y\right)\left(\frac{4}{9}x^{2}+\frac{1}{4}y^{2}-\frac{1}{3}xy\right)+2x+\left(\frac{1}{3}x+\frac{1}{2}y\right)\left(\frac{1}{3}x-\frac{1}{2}y\right)+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Use the distributive property to multiply \frac{2}{3}x+\frac{1}{2}y by \frac{4}{9}x^{2}+\frac{1}{4}y^{2}-\frac{1}{3}xy and combine like terms.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\left(\frac{8}{27}x^{3}+\frac{1}{8}y^{3}\right)+2x+\left(\frac{1}{3}x+\frac{1}{2}y\right)\left(\frac{1}{3}x-\frac{1}{2}y\right)+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

To find the opposite of \frac{8}{27}x^{3}+\frac{1}{8}y^{3}, find the opposite of each term.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\left(\frac{1}{3}x+\frac{1}{2}y\right)\left(\frac{1}{3}x-\frac{1}{2}y\right)+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Consider \left(\frac{1}{3}x+\frac{1}{2}y\right)\left(\frac{1}{3}x-\frac{1}{2}y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\left(\frac{1}{3}x\right)^{2}-\left(\frac{1}{2}y\right)^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

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

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\left(\frac{1}{3}\right)^{2}x^{2}-\left(\frac{1}{2}y\right)^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

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

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{1}{9}x^{2}-\left(\frac{1}{2}y\right)^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

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

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{1}{9}x^{2}-\left(\frac{1}{2}\right)^{2}y^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{1}{9}x^{2}-\frac{1}{4}y^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Combine \frac{4}{9}x^{2} and \frac{1}{9}x^{2} to get \frac{5}{9}x^{2}.

$$\frac{5}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x-\frac{1}{4}y^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Combine \frac{1}{4}y^{2} and -\frac{1}{4}y^{2} to get 0.

$$\frac{5}{9}x^{2}+\frac{2}{3}xy-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Use the distributive property to multiply \frac{2}{3}x by y-\frac{1}{6}x.

$$\frac{5}{9}x^{2}+\frac{2}{3}xy-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{2}{3}xy-\frac{1}{9}x^{2}$$

Combine \frac{2}{3}xy and \frac{2}{3}xy to get \frac{4}{3}xy.

$$\frac{5}{9}x^{2}+\frac{4}{3}xy-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x-\frac{1}{9}x^{2}$$

Combine \frac{5}{9}x^{2} and -\frac{1}{9}x^{2} to get \frac{4}{9}x^{2}.

$$\frac{4}{9}x^{2}+\frac{4}{3}xy-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x$$

Expand

$$-\frac{8x^{3}}{27}-\frac{y^{3}}{8}+\frac{4xy}{3}+\frac{4x^{2}}{9}+2x$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{2}{3}x+\frac{1}{2}y\right)^{2}.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\left(\frac{2}{3}x+\frac{1}{2}y\right)\left(\frac{4}{9}x^{2}+\frac{1}{4}y^{2}-\frac{1}{3}xy\right)+2x+\left(\frac{1}{3}x+\frac{1}{2}y\right)\left(\frac{1}{3}x-\frac{1}{2}y\right)+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Use the distributive property to multiply \frac{2}{3}x+\frac{1}{2}y by \frac{4}{9}x^{2}+\frac{1}{4}y^{2}-\frac{1}{3}xy and combine like terms.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\left(\frac{8}{27}x^{3}+\frac{1}{8}y^{3}\right)+2x+\left(\frac{1}{3}x+\frac{1}{2}y\right)\left(\frac{1}{3}x-\frac{1}{2}y\right)+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

To find the opposite of \frac{8}{27}x^{3}+\frac{1}{8}y^{3}, find the opposite of each term.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\left(\frac{1}{3}x+\frac{1}{2}y\right)\left(\frac{1}{3}x-\frac{1}{2}y\right)+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Consider \left(\frac{1}{3}x+\frac{1}{2}y\right)\left(\frac{1}{3}x-\frac{1}{2}y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\left(\frac{1}{3}x\right)^{2}-\left(\frac{1}{2}y\right)^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

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

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\left(\frac{1}{3}\right)^{2}x^{2}-\left(\frac{1}{2}y\right)^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

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

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{1}{9}x^{2}-\left(\frac{1}{2}y\right)^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

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

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{1}{9}x^{2}-\left(\frac{1}{2}\right)^{2}y^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{4}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{1}{9}x^{2}-\frac{1}{4}y^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Combine \frac{4}{9}x^{2} and \frac{1}{9}x^{2} to get \frac{5}{9}x^{2}.

$$\frac{5}{9}x^{2}+\frac{2}{3}xy+\frac{1}{4}y^{2}-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x-\frac{1}{4}y^{2}+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Combine \frac{1}{4}y^{2} and -\frac{1}{4}y^{2} to get 0.

$$\frac{5}{9}x^{2}+\frac{2}{3}xy-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{2}{3}x\left(y-\frac{1}{6}x\right)$$

Use the distributive property to multiply \frac{2}{3}x by y-\frac{1}{6}x.

$$\frac{5}{9}x^{2}+\frac{2}{3}xy-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x+\frac{2}{3}xy-\frac{1}{9}x^{2}$$

Combine \frac{2}{3}xy and \frac{2}{3}xy to get \frac{4}{3}xy.

$$\frac{5}{9}x^{2}+\frac{4}{3}xy-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x-\frac{1}{9}x^{2}$$

Combine \frac{5}{9}x^{2} and -\frac{1}{9}x^{2} to get \frac{4}{9}x^{2}.

$$\frac{4}{9}x^{2}+\frac{4}{3}xy-\frac{8}{27}x^{3}-\frac{1}{8}y^{3}+2x$$