Solve (sqrt{18}+3sqrt{3}-2sqrt{3})*2sqrt{3}

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

Evaluate

$$6\sqrt{6}+6\approx 20.696938457$$

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}.

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

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

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

Use the distributive property to multiply 2 by 3\sqrt{2}+\sqrt{3}.

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

Use the distributive property to multiply 6\sqrt{2}+2\sqrt{3} by \sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

$$6\sqrt{6}+2\times 3$$

Multiply 2 and 3 to get 6.

$$6\sqrt{6}+6$$