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

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

Evaluate

$$-4\sqrt{3}\approx -6.92820323$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

Subtract 4 from 4 to get 0.

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

Combine -2\sqrt{3} and -2\sqrt{3} to get -4\sqrt{3}.

$$-4\sqrt{3}$$

Expand

$$-4\sqrt{3}$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

Subtract 4 from 4 to get 0.

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

Combine -2\sqrt{3} and -2\sqrt{3} to get -4\sqrt{3}.

$$-4\sqrt{3}$$