Solve 1/3-sqrt{3}

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

Evaluate

$$\frac{\sqrt{3}}{6}+\frac{1}{2}\approx 0.788675135$$

Solution Steps

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

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

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

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

Square 3. Square \sqrt{3}.

$$\frac{3+\sqrt{3}}{9-3}$$

Subtract 3 from 9 to get 6.

$$\frac{3+\sqrt{3}}{6}$$

Factor

$$\frac{\sqrt{3} {(\sqrt{3} + 1)}}{6} = 0.7886751345948128$$