Solve sqrt{27}*sqrt{1/3}-(sqrt{7}+sqrt{3})(sqrt{7}-sqrt{3})

$$\sqrt { 27 } \times \sqrt { \frac { 1 } { 3 } } - ( \sqrt { 7 } + \sqrt { 3 } ) ( \sqrt { 7 } - \sqrt { 3 } )$$

Evaluate

$$-1$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$3\sqrt{3}\sqrt{\frac{1}{3}}-\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$3\sqrt{3}\times \frac{\sqrt{1}}{\sqrt{3}}-\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)$$

Calculate the square root of 1 and get 1.

$$3\sqrt{3}\times \frac{1}{\sqrt{3}}-\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)$$

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

$$3\sqrt{3}\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$3\sqrt{3}\times \frac{\sqrt{3}}{3}-\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)$$

Cancel out 3 and 3.

$$\sqrt{3}\sqrt{3}-\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)$$

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

$$3-\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)$$

Consider \left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\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}.

$$3-\left(\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$3-\left(7-\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$3-\left(7-3\right)$$

Subtract 3 from 7 to get 4.

$$3-4$$

Subtract 4 from 3 to get -1.

$$-1$$

Factor

$$-1$$