Can anyone solve this and explain it to me please. thank you!~!

Sqrt of 3 / 1 - Ceubroot of 2 + (Sqrt of 3)(Cuberoot of 2 )

2. Dear Anemori,

Is this what you mean, $\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$

3. Originally Posted by Sudharaka
Dear Anemori,

Is this what you mean, $\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$

Yes sir, that's what i meant.

4. Originally Posted by Sudharaka
Dear Anemori,

Is this what you mean, $\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$
Originally Posted by Anemori
Yes sir, that's what i meant.
Okay, then, what do you want to do with it? That is an "expression", not a "problem". Are you trying to add the two? To do that you will have to get fractions with the same denominator by multiplying $\displaystyle \sqrt{3}\sqrt[3]{2}$ by the unit fraction $\displaystyle \frac{1-\sqrt[3]{2}}{1-\sqrt[3]{2}}$.

5. Originally Posted by HallsofIvy
Okay, then, what do you want to do with it? That is an "expression", not a "problem". Are you trying to add the two? To do that you will have to get fractions with the same denominator by multiplying $\displaystyle \sqrt{3}\sqrt[3]{2}$ by the unit fraction $\displaystyle \frac{1-\sqrt[3]{2}}{1-\sqrt[3]{2}}$.

Simplified it. can you explain how do you do it? i understand you cannot multiply or divide both radicals with different index.

6. Dear Anemori,

$\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$

$\displaystyle \frac{\sqrt{3}+(1-\sqrt[3]{2})\sqrt{3}\sqrt[3]{2}}{1-\sqrt[3]{2}}$

$\displaystyle \frac{\sqrt{3}(1-\sqrt[3]{2}-\sqrt{3}(\sqrt[3]{2})^2)}{1-\sqrt[3]{2}}$

$\displaystyle \frac{\sqrt{3}(1-\sqrt[3]{2}-2^{\frac{2}{3}})}{1-\sqrt[3]{2}}$

7. Hello, Anemori!

Simplify: .$\displaystyle \frac{\sqrt{3}}{1 - \sqrt[3]{2}} + \sqrt{3}\cdot\sqrt[3]{2}$

Factor out $\displaystyle \sqrt{3}\!:\;\;\sqrt{3}\,\bigg[\frac{1}{1-\sqrt[3]{2}} + \sqrt[3]{2}\bigg]$

Multiply the fraction by $\displaystyle \frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{1 + \sqrt[3]{2} + \sqrt[3]{4}}$

. . $\displaystyle \frac{1}{1-\sqrt[3]{2}}\cdot\frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{1+\sqrt[3]{2} + \sqrt[3]{4}} \;=\; \frac{1+\sqrt[3]{2} + \sqrt[3]{4}}{1-2} \;=\;-\left(1 + \sqrt[3]{2} + \sqrt[3]{4}\right)$ .$\displaystyle =\;-1-\sqrt[3]{2} - \sqrt[3]{4}$

The problem becomes: . $\displaystyle \sqrt{3}\bigg[-1 - \sqrt[3]{2} - \sqrt[3]{4} + \sqrt[3]{2}\bigg] \;=\;\sqrt{3}\bigg[-1 - \sqrt[3]{4}\bigg]$

Therefore: .$\displaystyle -\sqrt{3}\left(1 + \sqrt[3]{4}\right)$