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

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

Evaluate

$$27$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Add 12 and 1 to get 13.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Add 13 and 4 to get 17.

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

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

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

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

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

Use the distributive property to multiply 4\sqrt{3}+2 by \sqrt{3}-1 and combine like terms.

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

The square of \sqrt{3} is 3.

$$17+2\sqrt{3}+4\times 3-2\sqrt{3}-2$$

Multiply 4 and 3 to get 12.

$$17+2\sqrt{3}+12-2\sqrt{3}-2$$

Subtract 2 from 12 to get 10.

$$17+2\sqrt{3}+10-2\sqrt{3}$$

Add 17 and 10 to get 27.

$$27+2\sqrt{3}-2\sqrt{3}$$

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

$$27$$

Factor

$$3^{3}$$