Solve left(sqrt{2}-sqrt{3}right)left(sqrt{2}+sqrt{3}right)+{left(sqrt{3}-2right)}^2+sqrt{48}

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

Evaluate

$$6$$

Solution Steps

Consider \left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

The square of \sqrt{2} is 2.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 2 to get -1.

$$-1+\left(\sqrt{3}-2\right)^{2}+\sqrt{48}$$

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

$$-1+\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4+\sqrt{48}$$

The square of \sqrt{3} is 3.

$$-1+3-4\sqrt{3}+4+\sqrt{48}$$

Add 3 and 4 to get 7.

$$-1+7-4\sqrt{3}+\sqrt{48}$$

Add -1 and 7 to get 6.

$$6-4\sqrt{3}+\sqrt{48}$$

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}.

$$6-4\sqrt{3}+4\sqrt{3}$$

Combine -4\sqrt{3} and 4\sqrt{3} to get 0.

$$6$$

Factor

$$2\times 3$$