Solve (2)-sqrt{36}+sqrt[3]{27}-sqrt{2}(sqrt{2}-1/sqrt{2})

$$( 2 ) - \sqrt { 36 } + \sqrt[ 3 ] { 27 } - \sqrt { 2 } ( \sqrt { 2 } - \frac { 1 } { \sqrt { 2 } } )$$

Evaluate

$$-2$$

Solution Steps

Calculate the square root of 36 and get 6.

$$2-6+\sqrt[3]{27}-\sqrt{2}\left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)$$

Subtract 6 from 2 to get -4.

$$-4+\sqrt[3]{27}-\sqrt{2}\left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)$$

Calculate \sqrt[3]{27} and get 3.

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

Add -4 and 3 to get -1.

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

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

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

The square of \sqrt{2} is 2.

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

Combine \sqrt{2} and -\frac{\sqrt{2}}{2} to get \frac{1}{2}\sqrt{2}.

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

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

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

Multiply 2 and \frac{1}{2} to get 1.

$$-1-1$$

Subtract 1 from -1 to get -2.

$$-2$$

Factor

$$-2$$