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

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

Evaluate

$$4$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

Add 1 and 2 to get 3.

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

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

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

The square of \sqrt{2} is 2.

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

Add 1 and 2 to get 3.

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

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

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

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

$$\left(9-4\left(\sqrt{2}\right)^{2}\right)\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)$$

The square of \sqrt{2} is 2.

$$\left(9-4\times 2\right)\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)$$

Multiply -4 and 2 to get -8.

$$\left(9-8\right)\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)$$

Subtract 8 from 9 to get 1.

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

Use the distributive property to multiply 1 by 4+2\sqrt{3}.

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

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

$$2^{2}$$