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

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

Evaluate

$$16-2\sqrt{2}\approx 13.171572875$$

Solution Steps

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

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

Expand \left(3\sqrt{2}\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

$$9\times 2-2^{2}+2-2\sqrt{2}$$

Multiply 9 and 2 to get 18.

$$18-2^{2}+2-2\sqrt{2}$$

Calculate 2 to the power of 2 and get 4.

$$18-4+2-2\sqrt{2}$$

Subtract 4 from 18 to get 14.

$$14+2-2\sqrt{2}$$

Add 14 and 2 to get 16.

$$16-2\sqrt{2}$$