Solve a^2+b^2=(a+b)^2+(a-b)^2/2

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

Solve for a

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

Solution Steps

Multiply both sides of the equation by 2.

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

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

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

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

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

Combine a^{2} and a^{2} to get 2a^{2}.

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

Combine 2ab and -2ab to get 0.

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

Combine b^{2} and b^{2} to get 2b^{2}.

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

Subtract 2a^{2} from both sides.

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

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

$$2b^{2}=2b^{2}$$

Cancel out 2 on both sides.

$$b^{2}=b^{2}$$

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 both sides of the equation by 2.

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

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

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

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

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

Combine a^{2} and a^{2} to get 2a^{2}.

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

Combine 2ab and -2ab to get 0.

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

Combine b^{2} and b^{2} to get 2b^{2}.

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

Subtract 2b^{2} from both sides.

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

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

$$2a^{2}=2a^{2}$$

Cancel out 2 on both sides.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

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