How to Rationalize the Denominator of a Radical Expression?

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To rationalize the denominator of the expression 1/(√[3]{a} + √[3]{b} + √[3]{c}), the discussion highlights using the identity for the sum of cubes, a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc). This leads to a transformation of the expression into a form that separates the radicals. Participants suggest using the difference of cubes formula, x^3 - y^3 = (x - y)(x^2 + xy + y^2), to further simplify the problem. The conversation includes playful exchanges about understanding and applying the mathematical concepts. Ultimately, the original poster confirms their solution with their teacher, validating the approach taken.
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Homework Statement


Rationalize the denominator..
<br /> \frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}<br />


Homework Equations



Algebraic equations.

The Attempt at a Solution


So using the form
a3 + b3 + c3 - 3abc= (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

So it becomes


<br /> \frac{\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2} -\sqrt[3]{ab}-\sqrt[3]{bc}-\sqrt[3]{ca} }{a + b + c - 3 \sqrt[3]{abc}}<br />


Actually I don't know what to do next
If i try (a+b)(a-b) then it doesn't work out ...
 
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Use

x^3-y^3=(x-y)(x^2+xy+y^2)

ehild
 


ehild said:
Use

x^3-y^3=(x-y)(x^2+xy+y^2)

ehild
I don't see how this applies to the OP's problem.
 


ehild said:
Use

x^3-y^3=(x-y)(x^2+xy+y^2)

ehild

Mark44 said:
I don't see how this applies to the OP's problem.

x-y=\frac{x^3-y^3}{x^2+xy+y^2}

:wink:
 


Good thing you put in the wink emoticon.
 


Of course! An emoticon is worth a thousands words.

If there were a more suitable emoticon for "I'm sure you get it now" I would use that one instead.

:smile: << (don't kill me for this :-p)
 


Yup got it!
Thanks!

(Checked it out with my teacher ... its right!)
 

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