Solve 3/16-3sqrt{3}

$$\frac { 3 } { 16 - 3 \sqrt { 3 } }$$

Evaluate

$$\frac{3\left(3\sqrt{3}+16\right)}{229}\approx 0.277678853$$

Solution Steps

Rationalize the denominator of \frac{3}{16-3\sqrt{3}} by multiplying numerator and denominator by 16+3\sqrt{3}.

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

Consider \left(16-3\sqrt{3}\right)\left(16+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}.

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

Calculate 16 to the power of 2 and get 256.

$$\frac{3\left(16+3\sqrt{3}\right)}{256-\left(-3\sqrt{3}\right)^{2}}$$

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

$$\frac{3\left(16+3\sqrt{3}\right)}{256-\left(-3\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$\frac{3\left(16+3\sqrt{3}\right)}{256-9\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3\left(16+3\sqrt{3}\right)}{256-9\times 3}$$

Multiply 9 and 3 to get 27.

$$\frac{3\left(16+3\sqrt{3}\right)}{256-27}$$

Subtract 27 from 256 to get 229.

$$\frac{3\left(16+3\sqrt{3}\right)}{229}$$

Use the distributive property to multiply 3 by 16+3\sqrt{3}.

$$\frac{48+9\sqrt{3}}{229}$$