Solve (sqrt{27}-3sqrt{2})^2+(sqrt{3}+2sqrt{2})^2

$$( \sqrt { 27 } - 3 \sqrt { 2 } ) ^ { 2 } + ( \sqrt { 3 } + 2 \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$56-14\sqrt{6}\approx 21.707143601$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 9 and 3 to get 27.

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

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

$$27-18\sqrt{6}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}+2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$27-18\sqrt{6}+9\times 2+\left(\sqrt{3}+2\sqrt{2}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$27-18\sqrt{6}+18+\left(\sqrt{3}+2\sqrt{2}\right)^{2}$$

Add 27 and 18 to get 45.

$$45-18\sqrt{6}+\left(\sqrt{3}+2\sqrt{2}\right)^{2}$$

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

$$45-18\sqrt{6}+\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$45-18\sqrt{6}+3+4\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

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

$$45-18\sqrt{6}+3+4\sqrt{6}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$45-18\sqrt{6}+3+4\sqrt{6}+4\times 2$$

Multiply 4 and 2 to get 8.

$$45-18\sqrt{6}+3+4\sqrt{6}+8$$

Add 3 and 8 to get 11.

$$45-18\sqrt{6}+11+4\sqrt{6}$$

Add 45 and 11 to get 56.

$$56-18\sqrt{6}+4\sqrt{6}$$

Combine -18\sqrt{6} and 4\sqrt{6} to get -14\sqrt{6}.

$$56-14\sqrt{6}$$

Expand

$$56-14\sqrt{6}$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 9 and 3 to get 27.

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

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

$$27-18\sqrt{6}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}+2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$27-18\sqrt{6}+9\times 2+\left(\sqrt{3}+2\sqrt{2}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$27-18\sqrt{6}+18+\left(\sqrt{3}+2\sqrt{2}\right)^{2}$$

Add 27 and 18 to get 45.

$$45-18\sqrt{6}+\left(\sqrt{3}+2\sqrt{2}\right)^{2}$$

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

$$45-18\sqrt{6}+\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$45-18\sqrt{6}+3+4\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}$$

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

$$45-18\sqrt{6}+3+4\sqrt{6}+4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$45-18\sqrt{6}+3+4\sqrt{6}+4\times 2$$

Multiply 4 and 2 to get 8.

$$45-18\sqrt{6}+3+4\sqrt{6}+8$$

Add 3 and 8 to get 11.

$$45-18\sqrt{6}+11+4\sqrt{6}$$

Add 45 and 11 to get 56.

$$56-18\sqrt{6}+4\sqrt{6}$$

Combine -18\sqrt{6} and 4\sqrt{6} to get -14\sqrt{6}.

$$56-14\sqrt{6}$$