Solve (2+sqrt{3})^2-(2+sqrt{3})(2-sqrt{3})+(2-sqrt{3})^2

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

Evaluate

$$13$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{3}\right)^{2}.

$$4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\left(2-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$4+4\sqrt{3}+3-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\left(2-\sqrt{3}\right)^{2}$$

Add 4 and 3 to get 7.

$$7+4\sqrt{3}-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+\left(2-\sqrt{3}\right)^{2}$$

Consider \left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$7+4\sqrt{3}-\left(4-\left(\sqrt{3}\right)^{2}\right)+\left(2-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$7+4\sqrt{3}-\left(4-3\right)+\left(2-\sqrt{3}\right)^{2}$$

Subtract 3 from 4 to get 1.

$$7+4\sqrt{3}-1+\left(2-\sqrt{3}\right)^{2}$$

Subtract 1 from 7 to get 6.

$$6+4\sqrt{3}+\left(2-\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-\sqrt{3}\right)^{2}.

$$6+4\sqrt{3}+4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$6+4\sqrt{3}+4-4\sqrt{3}+3$$

Add 4 and 3 to get 7.

$$6+4\sqrt{3}+7-4\sqrt{3}$$

Add 6 and 7 to get 13.

$$13+4\sqrt{3}-4\sqrt{3}$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$13$$

Factor

$$13$$