+205 Rationalize the denominator of $\displaystyle \frac{1}{\sqrt[3]{3} - \sqrt[3]{2}}$. With your answer in the form $\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}$, and the fraction in lowest terms, what is ?$A + B + C + D$
+591 We can write $\sqrt[3]{3}$ as $a$, and $\sqrt[3]{2}$ as $b$, and we know that
$(a-b)(a^2+b^2+ab)=a^3-b^3$, so
$a-b=\frac{a^3-b^3}{a^2+b^2+ab}=\sqrt[3]{3}-\sqrt[3]{2}=\frac1{\sqrt[3]{9}+\sqrt[3]{4}+\sqrt[3]{6}}$
that's only the denominator, so
$\frac1{\sqrt[3]{3}-\sqrt[3]{2}}=\frac1{\frac1{\sqrt[3]{9}+\sqrt[3]{4}+\sqrt[3]{6}}}=\frac{\sqrt[3]{\boxed{9}}+\sqrt[3]{\boxed{4}}+\sqrt[3]{\boxed{6}}}{\boxed1}$
so
A+B+C+D=9+4+6+1=$\boxed{20}$
