Solve (sqrt{2}-1)^2+(sqrt{2}+1)^2

$$( \sqrt { 2 } - 1 ) ^ { 2 } + ( \sqrt { 2 } + 1 ) ^ { 2 }$$

Evaluate

$$6$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1+\left(\sqrt{2}+1\right)^{2}$$

The square of \sqrt{2} is 2.

$$2-2\sqrt{2}+1+\left(\sqrt{2}+1\right)^{2}$$

Add 2 and 1 to get 3.

$$3-2\sqrt{2}+\left(\sqrt{2}+1\right)^{2}$$

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

$$3-2\sqrt{2}+\left(\sqrt{2}\right)^{2}+2\sqrt{2}+1$$

The square of \sqrt{2} is 2.

$$3-2\sqrt{2}+2+2\sqrt{2}+1$$

Add 2 and 1 to get 3.

$$3-2\sqrt{2}+3+2\sqrt{2}$$

Add 3 and 3 to get 6.

$$6-2\sqrt{2}+2\sqrt{2}$$

Combine -2\sqrt{2} and 2\sqrt{2} to get 0.

$$6$$

Factor

$$2\times 3$$