Solve (1/3)^-1-tan60^circ-(1+sqrt{2})^2+3/sqrt{3}

$$( \frac { 1 } { 3 } ) ^ { - 1 } - \tan 60 ^ { \circ } - ( 1 + \sqrt { 2 } ) ^ { 2 } + \frac { 3 } { \sqrt { 3 } }$$

Evaluate

$$-2\sqrt{2}\approx -2.828427125$$

Solution Steps

Calculate \frac{1}{3} to the power of -1 and get 3.

$$3-\tan(60)-\left(1+\sqrt{2}\right)^{2}+\frac{3}{\sqrt{3}}$$

Get the value of \tan(60) from trigonometric values table.

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

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

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

The square of \sqrt{2} is 2.

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

Add 1 and 2 to get 3.

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

To find the opposite of 3+2\sqrt{2}, find the opposite of each term.

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

Subtract 3 from 3 to get 0.

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

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

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

The square of \sqrt{3} is 3.

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

Cancel out 3 and 3.

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

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

$$-2\sqrt{2}$$