Solve (sqrt{3}+1)^2-(sqrt{2}+1)(sqrt{2}-1)+2(3-sqrt{3})

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

Evaluate

$$9$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Consider \left(\sqrt{2}+1\right)\left(\sqrt{2}-1\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 1.

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

The square of \sqrt{2} is 2.

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

Subtract 1 from 2 to get 1.

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

Subtract 1 from 4 to get 3.

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

Use the distributive property to multiply 2 by 3-\sqrt{3}.

$$3+2\sqrt{3}+6-2\sqrt{3}$$

Add 3 and 6 to get 9.

$$9+2\sqrt{3}-2\sqrt{3}$$

Combine 2\sqrt{3} and -2\sqrt{3} to get 0.

$$9$$

Factor

$$3^{2}$$