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

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

Evaluate

$$4$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Add 12 and 1 to get 13.

$$4-2\sqrt{3}+13-4\sqrt{3}+\left(\sqrt{3}-2\right)^{2}-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Add 4 and 13 to get 17.

$$17-2\sqrt{3}-4\sqrt{3}+\left(\sqrt{3}-2\right)^{2}-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Combine -2\sqrt{3} and -4\sqrt{3} to get -6\sqrt{3}.

$$17-6\sqrt{3}+\left(\sqrt{3}-2\right)^{2}-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

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

$$17-6\sqrt{3}+\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

The square of \sqrt{3} is 3.

$$17-6\sqrt{3}+3-4\sqrt{3}+4-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Add 3 and 4 to get 7.

$$17-6\sqrt{3}+7-4\sqrt{3}-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Add 17 and 7 to get 24.

$$24-6\sqrt{3}-4\sqrt{3}-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Combine -6\sqrt{3} and -4\sqrt{3} to get -10\sqrt{3}.

$$24-10\sqrt{3}-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Use the distributive property to multiply \sqrt{3}-1 by 2\sqrt{3}-1 and combine like terms.

$$24-10\sqrt{3}-\left(2\left(\sqrt{3}\right)^{2}-3\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

The square of \sqrt{3} is 3.

$$24-10\sqrt{3}-\left(2\times 3-3\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Multiply 2 and 3 to get 6.

$$24-10\sqrt{3}-\left(6-3\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Add 6 and 1 to get 7.

$$24-10\sqrt{3}-\left(7-3\sqrt{3}\right)-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

To find the opposite of 7-3\sqrt{3}, find the opposite of each term.

$$24-10\sqrt{3}-7+3\sqrt{3}-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Subtract 7 from 24 to get 17.

$$17-10\sqrt{3}+3\sqrt{3}-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Combine -10\sqrt{3} and 3\sqrt{3} to get -7\sqrt{3}.

$$17-7\sqrt{3}-\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Use the distributive property to multiply \sqrt{3}-1 by \sqrt{3}-2 and combine like terms.

$$17-7\sqrt{3}-\left(\left(\sqrt{3}\right)^{2}-3\sqrt{3}+2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

The square of \sqrt{3} is 3.

$$17-7\sqrt{3}-\left(3-3\sqrt{3}+2\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Add 3 and 2 to get 5.

$$17-7\sqrt{3}-\left(5-3\sqrt{3}\right)-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

To find the opposite of 5-3\sqrt{3}, find the opposite of each term.

$$17-7\sqrt{3}-5+3\sqrt{3}-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Subtract 5 from 17 to get 12.

$$12-7\sqrt{3}+3\sqrt{3}-\left(\sqrt{3}-1\right)\left(2\sqrt{3}-2\right)$$

Combine -7\sqrt{3} and 3\sqrt{3} to get -4\sqrt{3}.

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

Use the distributive property to multiply \sqrt{3}-1 by 2\sqrt{3}-2 and combine like terms.

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

The square of \sqrt{3} is 3.

$$12-4\sqrt{3}-\left(2\times 3-4\sqrt{3}+2\right)$$

Multiply 2 and 3 to get 6.

$$12-4\sqrt{3}-\left(6-4\sqrt{3}+2\right)$$

Add 6 and 2 to get 8.

$$12-4\sqrt{3}-\left(8-4\sqrt{3}\right)$$

To find the opposite of 8-4\sqrt{3}, find the opposite of each term.

$$12-4\sqrt{3}-8+4\sqrt{3}$$

Subtract 8 from 12 to get 4.

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

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

$$4$$

Factor

$$2^{2}$$