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

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

Evaluate

$$5y^{2}$$

Solution Steps

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

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

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

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

To find the opposite of \frac{1}{4}x^{2}+xy+y^{2}, find the opposite of each term.

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

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

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

Combine -xy and -xy to get -2xy.

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

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

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

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

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

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

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

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

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

To find the opposite of \frac{1}{4}x^{2}-y^{2}, find the opposite of each term.

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

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

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

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

$$-2xy+y^{2}+2xy+4y^{2}$$

Combine -2xy and 2xy to get 0.

$$y^{2}+4y^{2}$$

Combine y^{2} and 4y^{2} to get 5y^{2}.

$$5y^{2}$$

Expand

$$5y^{2}$$

Solution Steps

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

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

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

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

To find the opposite of \frac{1}{4}x^{2}+xy+y^{2}, find the opposite of each term.

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

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

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

Combine -xy and -xy to get -2xy.

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

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

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

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

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

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

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

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

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

To find the opposite of \frac{1}{4}x^{2}-y^{2}, find the opposite of each term.

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

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

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

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

$$-2xy+y^{2}+2xy+4y^{2}$$

Combine -2xy and 2xy to get 0.

$$y^{2}+4y^{2}$$

Combine y^{2} and 4y^{2} to get 5y^{2}.

$$5y^{2}$$