Solve (2/3x+1/3y)^2+(2/3x-1/3)^2+(frac{1}{3}x-frac{2}{3}y)(frac{1}{3}x+frac{2}{3}y)-4x(frac{1}{9}y-frac{1}{9})-(frac{1}{3})

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

Evaluate

$$-\frac{y^{2}}{3}+x^{2}-\frac{2}{9}$$

Solution Steps

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

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

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

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

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

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

Consider \left(\frac{1}{3}x-\frac{2}{3}y\right)\left(\frac{1}{3}x+\frac{2}{3}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}.

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

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

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

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

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

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

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

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

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

Combine \frac{8}{9}x^{2} and \frac{1}{9}x^{2} to get x^{2}.

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

Combine \frac{1}{9}y^{2} and -\frac{4}{9}y^{2} to get -\frac{1}{3}y^{2}.

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

Use the distributive property to multiply -4x by \frac{1}{9}y-\frac{1}{9}.

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

Combine \frac{4}{9}xy and -\frac{4}{9}xy to get 0.

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

Combine -\frac{4}{9}x and \frac{4}{9}x to get 0.

$$x^{2}-\frac{1}{3}y^{2}+\frac{1}{9}-\frac{1}{3}$$

Subtract \frac{1}{3} from \frac{1}{9} to get -\frac{2}{9}.

$$x^{2}-\frac{1}{3}y^{2}-\frac{2}{9}$$

Expand

$$-\frac{y^{2}}{3}+x^{2}-\frac{2}{9}$$

Solution Steps

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

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

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

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

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

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

Consider \left(\frac{1}{3}x-\frac{2}{3}y\right)\left(\frac{1}{3}x+\frac{2}{3}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}.

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

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

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

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

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

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

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

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

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

Combine \frac{8}{9}x^{2} and \frac{1}{9}x^{2} to get x^{2}.

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

Combine \frac{1}{9}y^{2} and -\frac{4}{9}y^{2} to get -\frac{1}{3}y^{2}.

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

Use the distributive property to multiply -4x by \frac{1}{9}y-\frac{1}{9}.

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

Combine \frac{4}{9}xy and -\frac{4}{9}xy to get 0.

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

Combine -\frac{4}{9}x and \frac{4}{9}x to get 0.

$$x^{2}-\frac{1}{3}y^{2}+\frac{1}{9}-\frac{1}{3}$$

Subtract \frac{1}{3} from \frac{1}{9} to get -\frac{2}{9}.

$$x^{2}-\frac{1}{3}y^{2}-\frac{2}{9}$$