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

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

Evaluate

$$9$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 4 to get 7.

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

Add 4 and 7 to get 11.

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

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

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

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

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

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

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

The square of \sqrt{3} is 3.

$$11+2\sqrt{3}-2\times 3-2\sqrt{3}+4$$

Multiply -2 and 3 to get -6.

$$11+2\sqrt{3}-6-2\sqrt{3}+4$$

Add -6 and 4 to get -2.

$$11+2\sqrt{3}-2-2\sqrt{3}$$

Subtract 2 from 11 to get 9.

$$9+2\sqrt{3}-2\sqrt{3}$$

Combine 2\sqrt{3} and -2\sqrt{3} to get 0.

$$9$$

Factor

$$3^{2}$$