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

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

Evaluate

$$6\sqrt{2}\approx 8.485281374$$

Solution Steps

The opposite of -3\sqrt{3} is 3\sqrt{3}.

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

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

$$\sqrt{9-18\sqrt{3}+9\left(\sqrt{3}\right)^{2}+\left(-3-3\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{9-18\sqrt{3}+9\times 3+\left(-3-3\sqrt{3}\right)^{2}}$$

Multiply 9 and 3 to get 27.

$$\sqrt{9-18\sqrt{3}+27+\left(-3-3\sqrt{3}\right)^{2}}$$

Add 9 and 27 to get 36.

$$\sqrt{36-18\sqrt{3}+\left(-3-3\sqrt{3}\right)^{2}}$$

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

$$\sqrt{36-18\sqrt{3}+9+18\sqrt{3}+9\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{36-18\sqrt{3}+9+18\sqrt{3}+9\times 3}$$

Multiply 9 and 3 to get 27.

$$\sqrt{36-18\sqrt{3}+9+18\sqrt{3}+27}$$

Add 9 and 27 to get 36.

$$\sqrt{36-18\sqrt{3}+36+18\sqrt{3}}$$

Add 36 and 36 to get 72.

$$\sqrt{72-18\sqrt{3}+18\sqrt{3}}$$

Combine -18\sqrt{3} and 18\sqrt{3} to get 0.

$$\sqrt{72}$$

Factor 72=6^{2}\times 2. Rewrite the square root of the product \sqrt{6^{2}\times 2} as the product of square roots \sqrt{6^{2}}\sqrt{2}. Take the square root of 6^{2}.

$$6\sqrt{2}$$