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

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

Evaluate

$$6\left(\sqrt{2}+3\right)\approx 26.485281374$$

Solution Steps

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

$$9\left(\sqrt{2}\right)^{2}+6\sqrt{2}+1-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)$$

The square of \sqrt{2} is 2.

$$9\times 2+6\sqrt{2}+1-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)$$

Multiply 9 and 2 to get 18.

$$18+6\sqrt{2}+1-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)$$

Add 18 and 1 to get 19.

$$19+6\sqrt{2}-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)$$

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.

$$19+6\sqrt{2}-\left(4-\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$19+6\sqrt{2}-\left(4-3\right)$$

Subtract 3 from 4 to get 1.

$$19+6\sqrt{2}-1$$

Subtract 1 from 19 to get 18.

$$18+6\sqrt{2}$$