Rationalizing a denominator involving the sum of 3 cube roots

[anemone]

Hi members of the forum,

Problem:

Rationalize the denominator of $\displaystyle \frac{1}{a^\frac{1}{3}+b^{\frac{1}{3}}+c^{\frac{1}{3}}}.$

I know that if we are asked to rationalize, say, something like $\displaystyle \frac{1}{1+2^{\frac{1}{3}}}$, what we could do is the following:

$\displaystyle \frac{1}{1+2^{\frac{1}{3}}}=\frac{1}{1+2^{\frac{1}{3}}}.\frac{1-2^{\frac{1}{3}}+2^{\frac{2}{3}}}{1-2^{\frac{1}{3}}+2^{\frac{2}{3}}}$

$\displaystyle \frac{1}{1+2^{\frac{1}{3}}}=\frac{1-2^{\frac{1}{3}}+2^{\frac{2}{3}}}{1-2^{\frac{1}{3}}+2^{\frac{2}{3}}+2^{\frac{1}{3}}-2^{\frac{2}{3}}+2}$

$\displaystyle \frac{1}{1+2^{\frac{1}{3}}}=\frac{1-2^{\frac{1}{3}}+2^{\frac{1}{3}}}{1+2}$

$\displaystyle \frac{1}{1+2^{\frac{1}{3}}}=\frac{1-2^{\frac{1}{3}}+2^{\frac{1}{3}}}{3}$

But this one seems like not to be the case with the problem that I am asking on this thread...

Could anyone give me some hints to tackle it?

Thanks in advance.

 

[Fernando Revilla]

Hi members of the forum,Rationalize the denominator of $\displaystyle \frac{1}{a^\frac{1}{3}+b^{\frac{1}{3}}+c^{\frac{1}{3}}}.$

We can use the equality
$$(x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) = x^3 + y^3 + z^3 - 3xyz$$
If $x=\sqrt[3]{a},y=\sqrt[3]{b},z=\sqrt[3]{c}$ and we multiply numerator and denominator by $x^2 + y^2 + z^2 - xy - xz - yz$ we get in the denominator $a+b+c-3\sqrt[3]{abc}$. Now, is easier to rationalize.

 

[anemone]

We can use the equality
$$(x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) = x^3 + y^3 + z^3 - 3xyz$$
If $x=\sqrt[3]{a},y=\sqrt[3]{b},z=\sqrt[3]{c}$ and we multiply numerator and denominator by $x^2 + y^2 + z^2 - xy - xz - yz$ we get in the denominator $a+b+c-3\sqrt[3]{abc}$. Now, is easier to rationalize.

Thanks, Fernando Revilla! While I am so happy to receive your help in such a short period of time, I can see it now that what you suggested to me is basically using the same principle that I used to tackle the example that I cited above, it's such a shame of me for not thinking harder before posting...(Headbang)

Guess that when I get stuck, the only place that I could think of (instead of keep plugging away) is MHB!:p

 

[MarkFL]

Many times the solution seems obvious...but only after it has been shown to us.(Nod)

And your posting has given others useful information...I wouldn't have known how to rationalize that denominator without furrowing my brows and committing pen to paper...(Yes)

 

[I like Serena]

Suppose you multiply numerator and denominator both with $a^3b^3c^3$.
Wouldn't that rationalize both your numerator and your denominator?

 

[MarkFL]

Suppose you multiply numerator and denominator both with $a^3b^3c^3$.
Wouldn't that rationalize both your numerator and your denominator?

The denominator would then be:

$\displaystyle a^{\frac{10}{3}}b^3c^3+a^3b^{\frac{10}{3}}c^3+a^3b^3c^{\frac{10}{3}}$

 

[I like Serena]

The denominator would then be:

$\displaystyle a^{\frac{10}{3}}b^3c^3+a^3b^{\frac{10}{3}}c^3+a^3b^3c^{\frac{10}{3}}$

Oops! (Blush)