Solve frac{3-3sqrt{3}}{4+3sqrt{3}}

$$\frac{ 3-3 \sqrt{ 3 } }{ 4+3 \sqrt{ 3 } }$$

Evaluate

$$\frac{21\sqrt{3}-39}{11}\approx -0.238812095$$

Solution Steps

Rationalize the denominator of \frac{3-3\sqrt{3}}{4+3\sqrt{3}} by multiplying numerator and denominator by 4-3\sqrt{3}.

$$\frac{\left(3-3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}{\left(4+3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}$$

Consider \left(4+3\sqrt{3}\right)\left(4-3\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}.

$$\frac{\left(3-3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}{4^{2}-\left(3\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(3-3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}{16-\left(3\sqrt{3}\right)^{2}}$$

Expand \left(3\sqrt{3}\right)^{2}.

$$\frac{\left(3-3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}{16-3^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(3-3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}{16-9\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(3-3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}{16-9\times 3}$$

Multiply 9 and 3 to get 27.

$$\frac{\left(3-3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}{16-27}$$

Subtract 27 from 16 to get -11.

$$\frac{\left(3-3\sqrt{3}\right)\left(4-3\sqrt{3}\right)}{-11}$$

Apply the distributive property by multiplying each term of 3-3\sqrt{3} by each term of 4-3\sqrt{3}.

$$\frac{12-9\sqrt{3}-12\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{-11}$$

Combine -9\sqrt{3} and -12\sqrt{3} to get -21\sqrt{3}.

$$\frac{12-21\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{-11}$$

The square of \sqrt{3} is 3.

$$\frac{12-21\sqrt{3}+9\times 3}{-11}$$

Multiply 9 and 3 to get 27.

$$\frac{12-21\sqrt{3}+27}{-11}$$

Add 12 and 27 to get 39.

$$\frac{39-21\sqrt{3}}{-11}$$

Factor

$$\frac{3 {(7 \sqrt{3} - 13)}}{11} = -0.23881209464123432$$