Solve {left(sqrt{3}-sqrt{2}right)}^2left(5+2sqrt{6}right)

$${ \left( \sqrt{ 3 } - \sqrt{ 2 } \right) }^{ 2 } \left( 5+2 \sqrt{ 6 } \right)$$

Evaluate

$$1$$

Solution Steps

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

$$\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)\left(5+2\sqrt{6}\right)$$

The square of \sqrt{3} is 3.

$$\left(3-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)\left(5+2\sqrt{6}\right)$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\left(3-2\sqrt{6}+\left(\sqrt{2}\right)^{2}\right)\left(5+2\sqrt{6}\right)$$

The square of \sqrt{2} is 2.

$$\left(3-2\sqrt{6}+2\right)\left(5+2\sqrt{6}\right)$$

Add 3 and 2 to get 5.

$$\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$25-\left(2\sqrt{6}\right)^{2}$$

Expand \left(2\sqrt{6}\right)^{2}.

$$25-2^{2}\left(\sqrt{6}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$25-4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$25-4\times 6$$

Multiply 4 and 6 to get 24.

$$25-24$$

Subtract 24 from 25 to get 1.

$$1$$

Factor

$$1$$