# How to simplify $\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$ to $\sqrt[3]{3}$

$$\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$$ 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.

• Just factor out a cube root of $3$ from your numerator, the denominator cancels against your term inside the brackets of the numerator Oct 21, 2015 at 15:30
• Try to use old trick with multiplying by one. Multiply it by $\frac{\sqrt[3]{9}-1}{\sqrt[3]{9}-1}$ Oct 21, 2015 at 15:30
• $a^3+b^3=(a+b)(a^2-ab+b^2)$, $a^3-b^3=(a-b)(a^2+ab+b^2)$
– user175968
Oct 21, 2015 at 15:34

Let $t=\sqrt[3]3$. Then, we have $t^3=3$, so
$$\frac{1}{t-1}-\frac{2}{t+1}=\frac{3-t}{t^2-1}=\frac{t(3-t)}{t(t^2-1)}=\frac{t(3-t)}{3-t}=t.$$
• @rop: If I didn't know the final result, I would have never tried to simplify it:). I just multiplied it by $t/t$ because the denominator has $t^2$. Oct 21, 2015 at 16:15
$$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}=\frac{(3^{1/3})^3-3^{1/3}}{3^{2/3}-1}=\frac{3^{1/3}(3^{2/3}-1)}{3^{2/3}-1}=3^{1/3}(=\sqrt[3]{3})$$