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

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

Evaluate

$$6\sqrt{3}+9\approx 19.392304845$$

Solution Steps

Rationalize the denominator of \frac{3\left(3+3\sqrt{3}\right)}{3-\sqrt{3}} by multiplying numerator and denominator by 3+\sqrt{3}.

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

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}.

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

Square 3. Square \sqrt{3}.

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

Subtract 3 from 9 to get 6.

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

Use the distributive property to multiply 3 by 3+3\sqrt{3}.

$$\frac{\left(9+9\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}$$

Apply the distributive property by multiplying each term of 9+9\sqrt{3} by each term of 3+\sqrt{3}.

$$\frac{27+9\sqrt{3}+27\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{6}$$

Combine 9\sqrt{3} and 27\sqrt{3} to get 36\sqrt{3}.

$$\frac{27+36\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{6}$$

The square of \sqrt{3} is 3.

$$\frac{27+36\sqrt{3}+9\times 3}{6}$$

Multiply 9 and 3 to get 27.

$$\frac{27+36\sqrt{3}+27}{6}$$

Add 27 and 27 to get 54.

$$\frac{54+36\sqrt{3}}{6}$$

Factor

$$3 \sqrt{3} {(\sqrt{3} + 2)} = 19.392304845$$