![How to simplify $\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$ to $\sqrt[3]{3}$](https://liberty.io/uploads/how-to-simplify-frac-1-sqrt-3-3-1-frac-2-sqrt-3-3-1-to-sqrt-3-3_720.jpg)
$\begingroup$$$\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$$ I have simplified above to: $$\frac{3-\sqrt[3]{3}}{(\sqrt[3]{3}+1)(\sqrt[3]{3}-1)}$$ What is equal to: $$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}$$ WolframAlpha says this can be shown as $\sqrt[3]{3}$, but I can't find out how to do this.
$\endgroup$32 Answers
$\begingroup$Let $t=\sqrt[3]3$. Then, we have $t^3=3$, so
$$\frac{1}{t-1}-\frac{2}{t+1}=\frac{3-t}{t^2-1}=\frac{t(3-t)}{t(t^2-1)}=\frac{t(3-t)}{3-t}=t.$$
$\endgroup$2$\begingroup$$$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}=\frac{(3^{1/3})^3-3^{1/3}}{3^{2/3}-1}=\frac{3^{1/3}(3^{2/3}-1)}{3^{2/3}-1}=3^{1/3}(=\sqrt[3]{3})$$
$\endgroup$1
