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

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

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}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$16-\left(2\sqrt{3}\right)^{2}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$16-2^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$16-4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$16-4\times 3$$

Multiply 4 and 3 to get 12.

$$16-12$$

Subtract 12 from 16 to get 4.

$$4$$

Factor

$$2^{2}$$