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

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

Evaluate

$$-2\sqrt{3}-15\approx -18.464101615$$

Solution Steps

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$1-12-\left(1+\sqrt{3}\right)^{2}$$

Subtract 12 from 1 to get -11.

$$-11-\left(1+\sqrt{3}\right)^{2}$$

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

$$-11-\left(1+2\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$-11-\left(1+2\sqrt{3}+3\right)$$

Add 1 and 3 to get 4.

$$-11-\left(4+2\sqrt{3}\right)$$

To find the opposite of 4+2\sqrt{3}, find the opposite of each term.

$$-11-4-2\sqrt{3}$$

Subtract 4 from -11 to get -15.

$$-15-2\sqrt{3}$$