Solve frac{sqrt[3]{27}+sqrt{27}-sqrt{48}}{2-sqrt{2}}

$$\frac { \sqrt[ 3 ] { 27 } + \sqrt { 27 } - \sqrt { 48 } } { 2 - \sqrt { 2 } }$$

Evaluate

$$\frac{3\sqrt{2}}{2}-\frac{\sqrt{6}}{2}+3-\sqrt{3}\approx 2.164524665$$

Solution Steps

Calculate \sqrt[3]{27} and get 3.

$$\frac{3+\sqrt{27}-\sqrt{48}}{2-\sqrt{2}}$$

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}.

$$\frac{3+3\sqrt{3}-\sqrt{48}}{2-\sqrt{2}}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

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

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

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

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

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

Consider \left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 2. Square \sqrt{2}.

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

Subtract 2 from 4 to get 2.

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

Use the distributive property to multiply 3-\sqrt{3} by 2+\sqrt{2}.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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