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

$$\left. \begin{array} { l } { ( \sqrt { 3 } + 2 ) ^ { 2 } - ( \sqrt { 3 } + 2 ) \times ( 2 - \sqrt { 3 } ) } \\ { + ( 2 - \sqrt { 3 } ) ^ { 2 } } \end{array} \right.$$

Evaluate

$$13$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 4 to get 7.

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

Consider \left(\sqrt{3}+2\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

$$13$$