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 )
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 $\displaystyle \frac{1-\sqrt[3]{2}}{1-\sqrt[3]{2}}$.
Dear Anemori,
$\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$
$\displaystyle \frac{\sqrt{3}+(1-\sqrt[3]{2})\sqrt{3}\sqrt[3]{2}}{1-\sqrt[3]{2}}$
$\displaystyle \frac{\sqrt{3}(1-\sqrt[3]{2}-\sqrt{3}(\sqrt[3]{2})^2)}{1-\sqrt[3]{2}}$
$\displaystyle \frac{\sqrt{3}(1-\sqrt[3]{2}-2^{\frac{2}{3}})}{1-\sqrt[3]{2}}$
Hope this solves the questions you have about radical expressions.
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) $