Solve (frac{2000sqrt{3}}{3-sqrt{3}})*3+sqrt{3}

$$( \frac{ 2000 \sqrt{ 3 } }{ 3- \sqrt{ 3 } } ) \times 3+ \sqrt{ 3 }$$

Evaluate

$$3001\sqrt{3}+3000\approx 8197.884473514$$

Short Solution Steps

Rationalize the denominator of \frac{2000\sqrt{3}}{3-\sqrt{3}} by multiplying numerator and denominator by 3+\sqrt{3}.

$$3\times \frac{2000\sqrt{3}\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\sqrt{3}$$

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}.

$$3\times \frac{2000\sqrt{3}\left(3+\sqrt{3}\right)}{3^{2}-\left(\sqrt{3}\right)^{2}}+\sqrt{3}$$

Square 3. Square \sqrt{3}.

$$3\times \frac{2000\sqrt{3}\left(3+\sqrt{3}\right)}{9-3}+\sqrt{3}$$

Subtract 3 from 9 to get 6.

$$3\times \frac{2000\sqrt{3}\left(3+\sqrt{3}\right)}{6}+\sqrt{3}$$

Cancel out 6, the greatest common factor in 3 and 6.

$$\frac{2000\sqrt{3}\left(3+\sqrt{3}\right)}{2}+\sqrt{3}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{3} times \frac{2}{2}.

$$\frac{2000\sqrt{3}\left(3+\sqrt{3}\right)}{2}+\frac{2\sqrt{3}}{2}$$

Since \frac{2000\sqrt{3}\left(3+\sqrt{3}\right)}{2} and \frac{2\sqrt{3}}{2} have the same denominator, add them by adding their numerators.

$$\frac{2000\sqrt{3}\left(3+\sqrt{3}\right)+2\sqrt{3}}{2}$$

Do the multiplications in 2000\sqrt{3}\left(3+\sqrt{3}\right)+2\sqrt{3}.

$$\frac{6000\sqrt{3}+6000+2\sqrt{3}}{2}$$

Do the calculations in 6000\sqrt{3}+6000+2\sqrt{3}.

$$\frac{6002\sqrt{3}+6000}{2}$$

Divide each term of 6002\sqrt{3}+6000 by 2 to get 3001\sqrt{3}+3000.

$$3001\sqrt{3}+3000$$

Factor

$$3001 \sqrt{3} + 3000 = 8197.884473514$$