Solve (sqrt{18}+sqrt{12})(3sqrt{2}-2sqrt{3})-(sqrt{3}-sqrt{2})^2

$$( \sqrt { 18 } + \sqrt { 12 } ) ( 3 \sqrt { 2 } - 2 \sqrt { 3 } ) - ( \sqrt { 3 } - \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$2\sqrt{6}+1\approx 5.898979486$$

Solution Steps

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

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

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Consider \left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-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}.

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

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

$$18-\left(2\sqrt{3}\right)^{2}-\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

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

$$18-2^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$18-4\left(\sqrt{3}\right)^{2}-\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$18-4\times 3-\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Multiply 4 and 3 to get 12.

$$18-12-\left(\sqrt{3}-\sqrt{2}\right)^{2}$$

Subtract 12 from 18 to get 6.

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

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

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

Add 3 and 2 to get 5.

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

To find the opposite of 5-2\sqrt{6}, find the opposite of each term.

$$6-5+2\sqrt{6}$$

Subtract 5 from 6 to get 1.

$$1+2\sqrt{6}$$