Solve (A-B)^2=(A-B)*(A-B)=

$$( A - B ) ^ { 2 } = ( A - B ) \times ( A - B ) =$$

Solve for A (complex solution)

$$A\in \mathrm{C}$$

Solution Steps

Multiply A-B and A-B to get \left(A-B\right)^{2}.

$$\left(A-B\right)^{2}=\left(A-B\right)^{2}$$

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

$$A^{2}-2AB+B^{2}=\left(A-B\right)^{2}$$

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

$$A^{2}-2AB+B^{2}=A^{2}-2AB+B^{2}$$

Subtract A^{2} from both sides.

$$A^{2}-2AB+B^{2}-A^{2}=-2AB+B^{2}$$

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

$$-2AB+B^{2}=-2AB+B^{2}$$

Add 2AB to both sides.

$$-2AB+B^{2}+2AB=B^{2}$$

Combine -2AB and 2AB to get 0.

$$B^{2}=B^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any A.

$$A\in \mathrm{C}$$

Solve for B (complex solution)

$$B\in \mathrm{C}$$

Solution Steps

Multiply A-B and A-B to get \left(A-B\right)^{2}.

$$\left(A-B\right)^{2}=\left(A-B\right)^{2}$$

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

$$A^{2}-2AB+B^{2}=\left(A-B\right)^{2}$$

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

$$A^{2}-2AB+B^{2}=A^{2}-2AB+B^{2}$$

Add 2AB to both sides.

$$A^{2}-2AB+B^{2}+2AB=A^{2}+B^{2}$$

Combine -2AB and 2AB to get 0.

$$A^{2}+B^{2}=A^{2}+B^{2}$$

Subtract B^{2} from both sides.

$$A^{2}+B^{2}-B^{2}=A^{2}$$

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

$$A^{2}=A^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any B.

$$B\in \mathrm{C}$$

Solve for A

$$A\in \mathrm{R}$$

Solution Steps

Multiply A-B and A-B to get \left(A-B\right)^{2}.

$$\left(A-B\right)^{2}=\left(A-B\right)^{2}$$

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

$$A^{2}-2AB+B^{2}=\left(A-B\right)^{2}$$

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

$$A^{2}-2AB+B^{2}=A^{2}-2AB+B^{2}$$

Subtract A^{2} from both sides.

$$A^{2}-2AB+B^{2}-A^{2}=-2AB+B^{2}$$

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

$$-2AB+B^{2}=-2AB+B^{2}$$

Add 2AB to both sides.

$$-2AB+B^{2}+2AB=B^{2}$$

Combine -2AB and 2AB to get 0.

$$B^{2}=B^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any A.

$$A\in \mathrm{R}$$

Solve for B

$$B\in \mathrm{R}$$

Solution Steps

Multiply A-B and A-B to get \left(A-B\right)^{2}.

$$\left(A-B\right)^{2}=\left(A-B\right)^{2}$$

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

$$A^{2}-2AB+B^{2}=\left(A-B\right)^{2}$$

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

$$A^{2}-2AB+B^{2}=A^{2}-2AB+B^{2}$$

Add 2AB to both sides.

$$A^{2}-2AB+B^{2}+2AB=A^{2}+B^{2}$$

Combine -2AB and 2AB to get 0.

$$A^{2}+B^{2}=A^{2}+B^{2}$$

Subtract B^{2} from both sides.

$$A^{2}+B^{2}-B^{2}=A^{2}$$

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

$$A^{2}=A^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any B.

$$B\in \mathrm{R}$$