Solve (sqrt{3}+sqrt{12}-sqrt{18})^2+(sqrt{3}-sqrt{12}+sqrt{48})^2

$$( \sqrt { 3 } + \sqrt { 12 } - \sqrt { 18 } ) ^ { 2 } + ( \sqrt { 3 } - \sqrt { 12 } + \sqrt { 48 } ) ^ { 2 }$$

Evaluate

$$72-18\sqrt{6}\approx 27.90918463$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\left(\sqrt{3}+2\sqrt{3}-\sqrt{18}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

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

$$\left(3\sqrt{3}-\sqrt{18}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

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}.

$$\left(3\sqrt{3}-3\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

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

$$9\left(\sqrt{3}\right)^{2}-18\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3-18\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27-18\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

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

$$27-18\sqrt{6}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

The square of \sqrt{2} is 2.

$$27-18\sqrt{6}+9\times 2+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$27-18\sqrt{6}+18+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

Add 27 and 18 to get 45.

$$45-18\sqrt{6}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$45-18\sqrt{6}+\left(\sqrt{3}-2\sqrt{3}+\sqrt{48}\right)^{2}$$

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

$$45-18\sqrt{6}+\left(-\sqrt{3}+\sqrt{48}\right)^{2}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

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

Combine -\sqrt{3} and 4\sqrt{3} to get 3\sqrt{3}.

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

Expand \left(3\sqrt{3}\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$45-18\sqrt{6}+9\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$45-18\sqrt{6}+9\times 3$$

Multiply 9 and 3 to get 27.

$$45-18\sqrt{6}+27$$

Add 45 and 27 to get 72.

$$72-18\sqrt{6}$$

Expand

$$72-18\sqrt{6}$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\left(\sqrt{3}+2\sqrt{3}-\sqrt{18}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

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

$$\left(3\sqrt{3}-\sqrt{18}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

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}.

$$\left(3\sqrt{3}-3\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

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

$$9\left(\sqrt{3}\right)^{2}-18\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3-18\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27-18\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

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

$$27-18\sqrt{6}+9\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

The square of \sqrt{2} is 2.

$$27-18\sqrt{6}+9\times 2+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

Multiply 9 and 2 to get 18.

$$27-18\sqrt{6}+18+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

Add 27 and 18 to get 45.

$$45-18\sqrt{6}+\left(\sqrt{3}-\sqrt{12}+\sqrt{48}\right)^{2}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$45-18\sqrt{6}+\left(\sqrt{3}-2\sqrt{3}+\sqrt{48}\right)^{2}$$

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

$$45-18\sqrt{6}+\left(-\sqrt{3}+\sqrt{48}\right)^{2}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

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

Combine -\sqrt{3} and 4\sqrt{3} to get 3\sqrt{3}.

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

Expand \left(3\sqrt{3}\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$45-18\sqrt{6}+9\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$45-18\sqrt{6}+9\times 3$$

Multiply 9 and 3 to get 27.

$$45-18\sqrt{6}+27$$

Add 45 and 27 to get 72.

$$72-18\sqrt{6}$$