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

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

Evaluate

$$2\sqrt{3}-15\approx -11.535898385$$

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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Subtract 12 from 1 to get -11.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 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.

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