Solve 4-(1-2sqrt{3})*(1+2sqrt{3})-(2sqrt{3}-1)^2=4sqrt{3}

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

Verify

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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$4-\left(1-12\right)-\left(2\sqrt{3}-1\right)^{2}=4\sqrt{3}$$

Subtract 12 from 1 to get -11.

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

The opposite of -11 is 11.

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

Add 4 and 11 to get 15.

$$15-\left(2\sqrt{3}-1\right)^{2}=4\sqrt{3}$$

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

$$15-\left(4\left(\sqrt{3}\right)^{2}-4\sqrt{3}+1\right)=4\sqrt{3}$$

The square of \sqrt{3} is 3.

$$15-\left(4\times 3-4\sqrt{3}+1\right)=4\sqrt{3}$$

Multiply 4 and 3 to get 12.

$$15-\left(12-4\sqrt{3}+1\right)=4\sqrt{3}$$

Add 12 and 1 to get 13.

$$15-\left(13-4\sqrt{3}\right)=4\sqrt{3}$$

To find the opposite of 13-4\sqrt{3}, find the opposite of each term.

$$15-13+4\sqrt{3}=4\sqrt{3}$$

Subtract 13 from 15 to get 2.

$$2+4\sqrt{3}=4\sqrt{3}$$

Subtract 4\sqrt{3} from both sides.

$$2+4\sqrt{3}-4\sqrt{3}=0$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$2=0$$

Compare 2 and 0.

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