Rationalizing the denominator involving more than one square root

[SwimmingGoat]

Here's my problems:
How might you "rationalize the denominator" if the expression is \frac{1}{2+7√2+5√3} or \frac{1}{\sqrt[3]{2}+1}?
I know that in typical problems where we rationalize the denominator, we simply have to multiply the denominator and numerator by the conjugate of the denominator. For example, if the denominator were 2+√5, I'd simply multiply the top and bottom by 2-√5.
I haven't been in any math classes for two years, and I'm taking my final math course, and I simply don't remember how to do things like this! I'm not looking for someone to solve the whole thing, but I need suggestions or ideas on how I can solve these problems... Thank you so much for your help in advance!

 

[mfb]

You can solve the first problem in two steps - rationalize away one root, simplify, then get rid of the other root.

The second problem needs a clever choice for an expansion.
What you need as result is something like $\sqrt[3]{2}^3+1$ (=3), and this can be written as $\sqrt[3]{2}^3+1 = (\sqrt[3]{2}+1) \cdot x$ where you have to find x. Expand the fraction with x, and you get a nice denominator.

 

[SwimmingGoat]

Thank you so much for your help! For the second one, I applied your idea with the fact that a^3+b^3=(a+b)(a^2-ab-b^2). I let a=1and b=\sqrt[3]{2}.

 

[ehild]

Thank you so much for your help! For the second one, I applied your idea with the fact that a^3+b^3=(a+b)(a^2-ab-b^2). I let a=1and b=\sqrt[3]{2}.

Correctly: a^3+b^3=(a+b)(a^2-ab+b^2)

ehild

 

[SwimmingGoat]

Whoops! Thanks for catching that.

 

[Chestermiller]

Here's my problems:
How might you "rationalize the denominator" if the expression is \frac{1}{2+7√2+5√3}?

Do what you usually do. First multiply numerator and denominator by 2+7√2-5√3. This will get rid of the square root of 3 in the denominator, and leave only a term with the square root of 2 in the denominator. Then multiply numerator and denominator by the conjugate of that.