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

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

Evaluate

$$32\sqrt{3}+71-24\sqrt{6}-42\sqrt{2}\approx 8.240902396$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

$$8-4\sqrt{3}+\left(2+\sqrt{3}\right)^{2}\left(\sqrt{6}-\sqrt{3}\right)^{2}$$

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

$$8-4\sqrt{3}+\left(4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)\left(\sqrt{6}-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$8-4\sqrt{3}+\left(4+4\sqrt{3}+3\right)\left(\sqrt{6}-\sqrt{3}\right)^{2}$$

Add 4 and 3 to get 7.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{3}\right)^{2}.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(\left(\sqrt{6}\right)^{2}-2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{6} is 6.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-2\sqrt{3}\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-2\times 3\sqrt{2}+\left(\sqrt{3}\right)^{2}\right)$$

Multiply -2 and 3 to get -6.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-6\sqrt{2}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-6\sqrt{2}+3\right)$$

Add 6 and 3 to get 9.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(9-6\sqrt{2}\right)$$

Use the distributive property to multiply 7+4\sqrt{3} by 9-6\sqrt{2}.

$$8-4\sqrt{3}+63-42\sqrt{2}+36\sqrt{3}-24\sqrt{3}\sqrt{2}$$

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

$$8-4\sqrt{3}+63-42\sqrt{2}+36\sqrt{3}-24\sqrt{6}$$

Add 8 and 63 to get 71.

$$71-4\sqrt{3}-42\sqrt{2}+36\sqrt{3}-24\sqrt{6}$$

Combine -4\sqrt{3} and 36\sqrt{3} to get 32\sqrt{3}.

$$71+32\sqrt{3}-42\sqrt{2}-24\sqrt{6}$$

Expand

$$32 \sqrt{3} + 71 - 24 \sqrt{6} - 42 \sqrt{2} = 8.240902396$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

$$8-4\sqrt{3}+\left(2+\sqrt{3}\right)^{2}\left(\sqrt{6}-\sqrt{3}\right)^{2}$$

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

$$8-4\sqrt{3}+\left(4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)\left(\sqrt{6}-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$8-4\sqrt{3}+\left(4+4\sqrt{3}+3\right)\left(\sqrt{6}-\sqrt{3}\right)^{2}$$

Add 4 and 3 to get 7.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{3}\right)^{2}.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(\left(\sqrt{6}\right)^{2}-2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{6} is 6.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-2\sqrt{6}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-2\sqrt{3}\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-2\times 3\sqrt{2}+\left(\sqrt{3}\right)^{2}\right)$$

Multiply -2 and 3 to get -6.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-6\sqrt{2}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(6-6\sqrt{2}+3\right)$$

Add 6 and 3 to get 9.

$$8-4\sqrt{3}+\left(7+4\sqrt{3}\right)\left(9-6\sqrt{2}\right)$$

Use the distributive property to multiply 7+4\sqrt{3} by 9-6\sqrt{2}.

$$8-4\sqrt{3}+63-42\sqrt{2}+36\sqrt{3}-24\sqrt{3}\sqrt{2}$$

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

$$8-4\sqrt{3}+63-42\sqrt{2}+36\sqrt{3}-24\sqrt{6}$$

Add 8 and 63 to get 71.

$$71-4\sqrt{3}-42\sqrt{2}+36\sqrt{3}-24\sqrt{6}$$

Combine -4\sqrt{3} and 36\sqrt{3} to get 32\sqrt{3}.

$$71+32\sqrt{3}-42\sqrt{2}-24\sqrt{6}$$