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

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

Evaluate

$$26$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Add 12 and 1 to get 13.

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

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

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

The square of \sqrt{3} is 3.

$$13-4\sqrt{3}+4\times 3+4\sqrt{3}+1$$

Multiply 4 and 3 to get 12.

$$13-4\sqrt{3}+12+4\sqrt{3}+1$$

Add 12 and 1 to get 13.

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

Add 13 and 13 to get 26.

$$26-4\sqrt{3}+4\sqrt{3}$$

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

$$26$$

Factor

$$2\times 13$$