Solve frac{sqrt{27}+sqrt{3}}{sqrt{3}}-frac{sqrt{6*sqrt{3}}}{sqrt{2}}

$$\frac{ \sqrt{ 27 } + \sqrt{ 3 } }{ \sqrt{ 3 } } - \frac{ \sqrt{ 6 \times \sqrt{ 3 } } }{ \sqrt{ 2 } }$$

Evaluate

$$4-3^{\frac{3}{4}}\approx 1.720492943$$

Short Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$\frac{3\sqrt{3}+\sqrt{3}}{\sqrt{3}}-\frac{\sqrt{6\sqrt{3}}}{\sqrt{2}}$$

Combine 3\sqrt{3} and \sqrt{3} to get 4\sqrt{3}.

$$\frac{4\sqrt{3}}{\sqrt{3}}-\frac{\sqrt{6\sqrt{3}}}{\sqrt{2}}$$

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

$$\frac{4\sqrt{3}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{6\sqrt{3}}}{\sqrt{2}}$$

The square of \sqrt{3} is 3.

$$\frac{4\sqrt{3}\sqrt{3}}{3}-\frac{\sqrt{6\sqrt{3}}}{\sqrt{2}}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{4\times 3}{3}-\frac{\sqrt{6\sqrt{3}}}{\sqrt{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{12}{3}-\frac{\sqrt{6\sqrt{3}}}{\sqrt{2}}$$

Divide 12 by 3 to get 4.

$$4-\frac{\sqrt{6\sqrt{3}}}{\sqrt{2}}$$

Rationalize the denominator of \frac{\sqrt{6\sqrt{3}}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$4-\frac{\sqrt{6\sqrt{3}}\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$4-\frac{\sqrt{6\sqrt{3}}\sqrt{2}}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{2}{2}.

$$\frac{4\times 2}{2}-\frac{\sqrt{6\sqrt{3}}\sqrt{2}}{2}$$

Since \frac{4\times 2}{2} and \frac{\sqrt{6\sqrt{3}}\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4\times 2-\sqrt{6\sqrt{3}}\sqrt{2}}{2}$$

Do the multiplications in 4\times 2-\sqrt{6\sqrt{3}}\sqrt{2}.

$$\frac{8-2\sqrt[4]{27}}{2}$$

Divide each term of 8-2\sqrt[4]{27} by 2 to get 4-\sqrt[4]{27}.

$$4-\sqrt[4]{27}$$