Solve (sqrt{3}-sqrt{7})(3sqrt{6}+sqrt{7})

$$( \sqrt { 3 } - \sqrt { 7 } ) ( 3 \sqrt { 6 } + \sqrt { 7 } )$$

Evaluate

$$\sqrt{21}+9\sqrt{2}-3\sqrt{42}-7\approx -9.131724339$$

Solution Steps

Apply the distributive property by multiplying each term of \sqrt{3}-\sqrt{7} by each term of 3\sqrt{6}+\sqrt{7}.

$$3\sqrt{3}\sqrt{6}+\sqrt{3}\sqrt{7}-3\sqrt{7}\sqrt{6}-\left(\sqrt{7}\right)^{2}$$

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}.

$$3\sqrt{3}\sqrt{3}\sqrt{2}+\sqrt{3}\sqrt{7}-3\sqrt{7}\sqrt{6}-\left(\sqrt{7}\right)^{2}$$

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

$$3\times 3\sqrt{2}+\sqrt{3}\sqrt{7}-3\sqrt{7}\sqrt{6}-\left(\sqrt{7}\right)^{2}$$

Multiply 3 and 3 to get 9.

$$9\sqrt{2}+\sqrt{3}\sqrt{7}-3\sqrt{7}\sqrt{6}-\left(\sqrt{7}\right)^{2}$$

To multiply \sqrt{3} and \sqrt{7}, multiply the numbers under the square root.

$$9\sqrt{2}+\sqrt{21}-3\sqrt{7}\sqrt{6}-\left(\sqrt{7}\right)^{2}$$

To multiply \sqrt{7} and \sqrt{6}, multiply the numbers under the square root.

$$9\sqrt{2}+\sqrt{21}-3\sqrt{42}-\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{7} is 7.

$$9\sqrt{2}+\sqrt{21}-3\sqrt{42}-7$$