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

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

Evaluate

$$12\sqrt{2}\approx 16.970562748$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply 2 and 3 to get 6.

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

The square of \sqrt{3} is 3.

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

Add 6 and 3 to get 9.

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

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 3 to get -6.

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

The square of \sqrt{3} is 3.

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

Add 6 and 3 to get 9.

$$9+6\sqrt{2}-\left(9-6\sqrt{2}\right)$$

To find the opposite of 9-6\sqrt{2}, find the opposite of each term.

$$9+6\sqrt{2}-9+6\sqrt{2}$$

Subtract 9 from 9 to get 0.

$$6\sqrt{2}+6\sqrt{2}$$

Combine 6\sqrt{2} and 6\sqrt{2} to get 12\sqrt{2}.

$$12\sqrt{2}$$

Expand

$$12 \sqrt{2} = 16.970562748$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply 2 and 3 to get 6.

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

The square of \sqrt{3} is 3.

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

Add 6 and 3 to get 9.

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

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 3 to get -6.

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

The square of \sqrt{3} is 3.

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

Add 6 and 3 to get 9.

$$9+6\sqrt{2}-\left(9-6\sqrt{2}\right)$$

To find the opposite of 9-6\sqrt{2}, find the opposite of each term.

$$9+6\sqrt{2}-9+6\sqrt{2}$$

Subtract 9 from 9 to get 0.

$$6\sqrt{2}+6\sqrt{2}$$

Combine 6\sqrt{2} and 6\sqrt{2} to get 12\sqrt{2}.

$$12\sqrt{2}$$