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

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

Evaluate

$$2-4\sqrt{3}\approx -4.92820323$$

Solution Steps

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Subtract 12 from 1 to get -11.

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

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

$$-11+4\left(\sqrt{3}\right)^{2}-4\sqrt{3}+1$$

The square of \sqrt{3} is 3.

$$-11+4\times 3-4\sqrt{3}+1$$

Multiply 4 and 3 to get 12.

$$-11+12-4\sqrt{3}+1$$

Add 12 and 1 to get 13.

$$-11+13-4\sqrt{3}$$

Add -11 and 13 to get 2.

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