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

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

Evaluate

$$2\sqrt{3}+3\sqrt{2}-\sqrt{6}-3\approx 2.257252559$$

Solution Steps

Calculate \sqrt{6}-\sqrt{3} to the power of 1 and get \sqrt{6}-\sqrt{3}.

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

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

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

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}.

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

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

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

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}.

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

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

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

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

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

The square of \sqrt{3} is 3.

$$2\sqrt{3}+3\sqrt{2}-\sqrt{6}-3$$