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

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

Evaluate

$$\sqrt{2}+\sqrt{3}\approx 3.14626437$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

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

Add 3 and 2 to get 5.

$$\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)$$

Use the distributive property to multiply 5+2\sqrt{6} by \sqrt{3}-\sqrt{2}.

$$5\sqrt{3}-5\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{2}$$

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}.

$$5\sqrt{3}-5\sqrt{2}+2\sqrt{3}\sqrt{2}\sqrt{3}-2\sqrt{6}\sqrt{2}$$

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

$$5\sqrt{3}-5\sqrt{2}+2\times 3\sqrt{2}-2\sqrt{6}\sqrt{2}$$

Multiply 2 and 3 to get 6.

$$5\sqrt{3}-5\sqrt{2}+6\sqrt{2}-2\sqrt{6}\sqrt{2}$$

Combine -5\sqrt{2} and 6\sqrt{2} to get \sqrt{2}.

$$5\sqrt{3}+\sqrt{2}-2\sqrt{6}\sqrt{2}$$

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}.

$$5\sqrt{3}+\sqrt{2}-2\sqrt{2}\sqrt{3}\sqrt{2}$$

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

$$5\sqrt{3}+\sqrt{2}-2\times 2\sqrt{3}$$

Multiply -2 and 2 to get -4.

$$5\sqrt{3}+\sqrt{2}-4\sqrt{3}$$

Combine 5\sqrt{3} and -4\sqrt{3} to get \sqrt{3}.

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