Solve sqrt[6]{3}-(sqrt{3}+1)^2

$$\sqrt[ 6 ] { 3 } - ( \sqrt { 3 } + 1 ) ^ { 2 }$$

Evaluate

$$\sqrt[6]{3}-2\sqrt{3}-4\approx -6.26316466$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

$$\sqrt[6]{3}-\left(4+2\sqrt{3}\right)$$

To find the opposite of 4+2\sqrt{3}, find the opposite of each term.

$$\sqrt[6]{3}-4-2\sqrt{3}$$