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

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

Evaluate

$$21$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

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

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

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

The square of \sqrt{3} is 3.

$$18-6\sqrt{6}+3+6\sqrt{6}$$

Add 18 and 3 to get 21.

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

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

$$21$$

Factor

$$3\times 7$$