Solve (sqrt{3}+3sqrt{2}-sqrt{6})(sqrt{3}-3sqrt{2}-sqrt{6})

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

Evaluate

$$-6\sqrt{2}-9\approx -17.485281374$$

Solution Steps

Apply the distributive property by multiplying each term of \sqrt{3}+3\sqrt{2}-\sqrt{6} by each term of \sqrt{3}-3\sqrt{2}-\sqrt{6}.

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

The square of \sqrt{3} is 3.

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

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

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

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

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

Combine -3\sqrt{6} and 3\sqrt{6} to get 0.

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

The square of \sqrt{2} is 2.

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

Multiply -9 and 2 to get -18.

$$3-3\sqrt{2}-18-3\sqrt{2}\sqrt{6}-\sqrt{6}\sqrt{3}+3\sqrt{6}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Subtract 18 from 3 to get -15.

$$-15-3\sqrt{2}-3\sqrt{2}\sqrt{6}-\sqrt{6}\sqrt{3}+3\sqrt{6}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$-15-3\sqrt{2}-3\sqrt{2}\sqrt{2}\sqrt{3}-\sqrt{6}\sqrt{3}+3\sqrt{6}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$-15-3\sqrt{2}-3\times 2\sqrt{3}-\sqrt{6}\sqrt{3}+3\sqrt{6}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Multiply -3 and 2 to get -6.

$$-15-3\sqrt{2}-6\sqrt{3}-\sqrt{6}\sqrt{3}+3\sqrt{6}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$-15-3\sqrt{2}-6\sqrt{3}-\sqrt{3}\sqrt{2}\sqrt{3}+3\sqrt{6}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$-15-3\sqrt{2}-6\sqrt{3}-3\sqrt{2}+3\sqrt{6}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Combine -3\sqrt{2} and -3\sqrt{2} to get -6\sqrt{2}.

$$-15-6\sqrt{2}-6\sqrt{3}+3\sqrt{6}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$-15-6\sqrt{2}-6\sqrt{3}+3\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$-15-6\sqrt{2}-6\sqrt{3}+3\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

Multiply 3 and 2 to get 6.

$$-15-6\sqrt{2}-6\sqrt{3}+6\sqrt{3}+\left(\sqrt{6}\right)^{2}$$

Combine -6\sqrt{3} and 6\sqrt{3} to get 0.

$$-15-6\sqrt{2}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$-15-6\sqrt{2}+6$$

Add -15 and 6 to get -9.

$$-9-6\sqrt{2}$$