Solve 3sqrt{3}(7sqrt{21}+sqrt{6})

$$3 \sqrt{ 3 } (7 \sqrt{ 21 } + \sqrt{ 6 } )$$

Evaluate

$$9\left(\sqrt{2}+7\sqrt{7}\right)\approx 179.410254658$$

Solution Steps

Use the distributive property to multiply 3\sqrt{3} by 7\sqrt{21}+\sqrt{6}.

$$21\sqrt{3}\sqrt{21}+3\sqrt{3}\sqrt{6}$$

Factor 21=3\times 7. Rewrite the square root of the product \sqrt{3\times 7} as the product of square roots \sqrt{3}\sqrt{7}.

$$21\sqrt{3}\sqrt{3}\sqrt{7}+3\sqrt{3}\sqrt{6}$$

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

$$21\times 3\sqrt{7}+3\sqrt{3}\sqrt{6}$$

Multiply 21 and 3 to get 63.

$$63\sqrt{7}+3\sqrt{3}\sqrt{6}$$

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}.

$$63\sqrt{7}+3\sqrt{3}\sqrt{3}\sqrt{2}$$

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

$$63\sqrt{7}+3\times 3\sqrt{2}$$

Multiply 3 and 3 to get 9.

$$63\sqrt{7}+9\sqrt{2}$$