Solve (sqrt{3}+1)(sqrt{3}-1)-sqrt{3}*sqrt{1/18}

$$( \sqrt { 3 } + 1 ) ( \sqrt { 3 } - 1 ) - \sqrt { 3 } \times \sqrt { \frac { 1 } { 18 } }$$

Evaluate

$$-\frac{\sqrt{6}}{6}+2\approx 1.59175171$$

Short Solution Steps

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-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}.

$$\left(\sqrt{3}\right)^{2}-1^{2}-\sqrt{3}\sqrt{\frac{1}{18}}$$

The square of \sqrt{3} is 3.

$$3-1^{2}-\sqrt{3}\sqrt{\frac{1}{18}}$$

Calculate 1 to the power of 2 and get 1.

$$3-1-\sqrt{3}\sqrt{\frac{1}{18}}$$

Subtract 1 from 3 to get 2.

$$2-\sqrt{3}\sqrt{\frac{1}{18}}$$

Rewrite the square root of the division \sqrt{\frac{1}{18}} as the division of square roots \frac{\sqrt{1}}{\sqrt{18}}.

$$2-\sqrt{3}\times \frac{\sqrt{1}}{\sqrt{18}}$$

Calculate the square root of 1 and get 1.

$$2-\sqrt{3}\times \frac{1}{\sqrt{18}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$2-\sqrt{3}\times \frac{1}{3\sqrt{2}}$$

Rationalize the denominator of \frac{1}{3\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

$$2-\sqrt{3}\times \frac{\sqrt{2}}{3\times 2}$$

Multiply 3 and 2 to get 6.

$$2-\sqrt{3}\times \frac{\sqrt{2}}{6}$$

Express \sqrt{3}\times \frac{\sqrt{2}}{6} as a single fraction.

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

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

$$2-\frac{\sqrt{6}}{6}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{6}{6}.

$$\frac{2\times 6}{6}-\frac{\sqrt{6}}{6}$$

Since \frac{2\times 6}{6} and \frac{\sqrt{6}}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\times 6-\sqrt{6}}{6}$$

Do the multiplications in 2\times 6-\sqrt{6}.

$$\frac{12-\sqrt{6}}{6}$$