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

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

Evaluate

$$0$$

Solution Steps

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

$$\sqrt{4\left(\sqrt{3}\right)^{2}-12\sqrt{3}+9}-\sqrt{\left(3-2\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{4\times 3-12\sqrt{3}+9}-\sqrt{\left(3-2\sqrt{3}\right)^{2}}$$

Multiply 4 and 3 to get 12.

$$\sqrt{12-12\sqrt{3}+9}-\sqrt{\left(3-2\sqrt{3}\right)^{2}}$$

Add 12 and 9 to get 21.

$$\sqrt{21-12\sqrt{3}}-\sqrt{\left(3-2\sqrt{3}\right)^{2}}$$

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

$$\sqrt{21-12\sqrt{3}}-\sqrt{9-12\sqrt{3}+4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{21-12\sqrt{3}}-\sqrt{9-12\sqrt{3}+4\times 3}$$

Multiply 4 and 3 to get 12.

$$\sqrt{21-12\sqrt{3}}-\sqrt{9-12\sqrt{3}+12}$$

Add 9 and 12 to get 21.

$$\sqrt{21-12\sqrt{3}}-\sqrt{21-12\sqrt{3}}$$

Combine \sqrt{21-12\sqrt{3}} and -\sqrt{21-12\sqrt{3}} to get 0.

$$0$$

Factor

$$0$$