### Instruction: Rationalize the denominator: $$\frac{1}{\sqrt[3]{2}+\sqrt[3]{16}}$$.### Response: To rationalize the denominator of a fraction, I want to multiply the fraction by something that will eliminate the cube roots from the bottom.
One way to do this is to use the fact that $a^3+b^3=(a+b)(a^2-ab+b^2)$.
If I let $a=\sqrt[3]{2}$ and $b=\sqrt[3]{16}$, then $a^3+b^3=2+16=18$.
So, I can multiply the fraction by $\frac{a^2-ab+b^2}{a^2-ab+b^2}$, which is equivalent to 1, and get: $$\frac{1}{\sqrt[3]{2}+\sqrt[3]{16}}\cdot\frac{a^2-ab+b^2}{a^2-ab+b^2}=\frac{a^2-ab+b^2}{18}$$
Now, I just need to simplify the numerator by plugging in the values of $a$ and $b$: $$\frac{a^2-ab+b^2}{18}=\frac{(\sqrt[3]{2})^2-(\sqrt[3]{2})(\sqrt[3]{16})+(\sqrt[3]{16})^2}{18}$$
To simplify the cube roots, I can use the property that $(a^m)^n=a^{mn}$: $$\frac{(\sqrt[3]{2})^2-(\sqrt[3]{2})(\sqrt[3]{16})+(\sqrt[3]{16})^2}{18}=\frac{2^{2/3}-2^{1/3}\cdot 2^{4/3}+2^{8/3}}{18}$$
Then, I can combine the exponents of 2 using the rules of exponentiation: $$\frac{2^{2/3}-2^{1/3}\cdot 2^{4/3}+2^{8/3}}{18}=\frac{2^{2/3}-2^{5/3}+2^{8/3}}{18}$$
Finally, I can factor out a common factor of $2^{2/3}$ from the numerator: $$\frac{2^{2/3}-2^{5/3}+2^{8/3}}{18}=\frac{2^{2/3}(1-2+2^2)}{18}$$
This simplifies to: $$\frac{2^{2/3}(1-2+2^2)}{18}=\frac{2^{2/3}\cdot 3}{18}=\frac{2^{2/3}}{6}$$