Solve (x-y)^2=x^2-2xy+y^2

$$( x - y ) ^ { 2 } = x ^ { 2 } - 2 x y + y ^ { 2 }$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

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

$$x^{2}-2xy+y^{2}=x^{2}-2xy+y^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-2xy+y^{2}-x^{2}=-2xy+y^{2}$$

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

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

Add 2xy to both sides.

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

Combine -2xy and 2xy to get 0.

$$y^{2}=y^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for y (complex solution)

$$y\in \mathrm{C}$$

Solution Steps

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

$$x^{2}-2xy+y^{2}=x^{2}-2xy+y^{2}$$

Add 2xy to both sides.

$$x^{2}-2xy+y^{2}+2xy=x^{2}+y^{2}$$

Combine -2xy and 2xy to get 0.

$$x^{2}+y^{2}=x^{2}+y^{2}$$

Subtract y^{2} from both sides.

$$x^{2}+y^{2}-y^{2}=x^{2}$$

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

$$x^{2}=x^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any y.

$$y\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

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

$$x^{2}-2xy+y^{2}=x^{2}-2xy+y^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-2xy+y^{2}-x^{2}=-2xy+y^{2}$$

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

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

Add 2xy to both sides.

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

Combine -2xy and 2xy to get 0.

$$y^{2}=y^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$

Solve for y

$$y\in \mathrm{R}$$

Solution Steps

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

$$x^{2}-2xy+y^{2}=x^{2}-2xy+y^{2}$$

Add 2xy to both sides.

$$x^{2}-2xy+y^{2}+2xy=x^{2}+y^{2}$$

Combine -2xy and 2xy to get 0.

$$x^{2}+y^{2}=x^{2}+y^{2}$$

Subtract y^{2} from both sides.

$$x^{2}+y^{2}-y^{2}=x^{2}$$

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

$$x^{2}=x^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any y.

$$y\in \mathrm{R}$$