Solve 2+[(2sqrt{3}-3sqrt{2})^2:6]*(5+2sqrt{6})

$$2 + [ ( 2 \sqrt { 3 } - 3 \sqrt { 2 } ) ^ { 2 } : 6 ] \cdot ( 5 + 2 \sqrt { 6 } )$$

Evaluate

$$3$$

Solution Steps

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

$$2+\frac{4\left(\sqrt{3}\right)^{2}-12\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}}{6}\left(5+2\sqrt{6}\right)$$

The square of \sqrt{3} is 3.

$$2+\frac{4\times 3-12\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}}{6}\left(5+2\sqrt{6}\right)$$

Multiply 4 and 3 to get 12.

$$2+\frac{12-12\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}}{6}\left(5+2\sqrt{6}\right)$$

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

$$2+\frac{12-12\sqrt{6}+9\left(\sqrt{2}\right)^{2}}{6}\left(5+2\sqrt{6}\right)$$

The square of \sqrt{2} is 2.

$$2+\frac{12-12\sqrt{6}+9\times 2}{6}\left(5+2\sqrt{6}\right)$$

Multiply 9 and 2 to get 18.

$$2+\frac{12-12\sqrt{6}+18}{6}\left(5+2\sqrt{6}\right)$$

Add 12 and 18 to get 30.

$$2+\frac{30-12\sqrt{6}}{6}\left(5+2\sqrt{6}\right)$$

Divide each term of 30-12\sqrt{6} by 6 to get 5-2\sqrt{6}.

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

Consider \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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{6} is 6.

$$2+25-4\times 6$$

Multiply 4 and 6 to get 24.

$$2+25-24$$

Subtract 24 from 25 to get 1.

$$2+1$$

Add 2 and 1 to get 3.

$$3$$

Factor

$$3$$