Solve frac{sqrt{2}*sqrt{6}}{sqrt{3}}+(sqrt{3}-1)^2-(sqrt{2}+1)(sqrt{2}-1)

$$\frac { \sqrt { 2 } \times \sqrt { 6 } } { \sqrt { 3 } } + ( \sqrt { 3 } - 1 ) ^ { 2 } - ( \sqrt { 2 } + 1 ) ( \sqrt { 2 } - 1 )$$

Evaluate

$$5-2\sqrt{3}\approx 1.535898385$$

Solution Steps

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}.

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

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

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

Rationalize the denominator of \frac{2\sqrt{3}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

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

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

Multiply 2 and 3 to get 6.

$$\frac{6}{3}+\left(\sqrt{3}-1\right)^{2}-\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)$$

Divide 6 by 3 to get 2.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Add 2 and 4 to get 6.

$$6-2\sqrt{3}-\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)$$

Consider \left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$6-2\sqrt{3}-\left(\left(\sqrt{2}\right)^{2}-1\right)$$

The square of \sqrt{2} is 2.

$$6-2\sqrt{3}-\left(2-1\right)$$

Subtract 1 from 2 to get 1.

$$6-2\sqrt{3}-1$$

Subtract 1 from 6 to get 5.

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