Solve (a-b)^2=(a+b)^2-4ab

$$( a - b ) ^ { 2 } = ( a + b ) ^ { 2 } - 4 a b$$

Solve for a

$$a\in \mathrm{R}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$a^{2}-2ab+b^{2}=\left(a+b\right)^{2}-4ab$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}-2ab+b^{2}=a^{2}+2ab+b^{2}-4ab$$

Combine 2ab and -4ab to get -2ab.

$$a^{2}-2ab+b^{2}=a^{2}-2ab+b^{2}$$

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$a^{2}-2ab+b^{2}=\left(a+b\right)^{2}-4ab$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

$$a^{2}-2ab+b^{2}=a^{2}+2ab+b^{2}-4ab$$

Combine 2ab and -4ab to get -2ab.

$$a^{2}-2ab+b^{2}=a^{2}-2ab+b^{2}$$

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.

$$b\in \mathrm{R}$$