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It is a very difficult question. How can we Rationalizing the denominator? $$\frac{2^{1/2}}{5+3*(4^{1/3})-7*(2^{1/3})}$$

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  • $\begingroup$ Expand the fraction by $(67 + 53\cdot 2^{1/3} + 34\cdot 4^{1/3})$. $\endgroup$ – Arthur Feb 19 '14 at 9:05
  • $\begingroup$ Hint: Use complex numbers. $\endgroup$ – user2369284 Feb 19 '14 at 9:06
  • $\begingroup$ I'd hate to think you might have to factor that denominator first. It doesn't look clean... $\endgroup$ – Mike Feb 19 '14 at 9:07
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    $\begingroup$ @Mike It's a polynomial in $2^{1/3}$, and you cannot factor $5 - 7x + 3x^2$ in any nice way (without using complex numbers). So it's as clean as it's going to get. $\endgroup$ – Arthur Feb 19 '14 at 9:09
  • $\begingroup$ @Arthur Agreed. But it would be easier to rationalize terms of $a+b\times2^{1/3}$. It would be a ******* mess, but it would be straightforward. Where does your number come from? $\endgroup$ – Mike Feb 19 '14 at 9:19
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Your denominator is a polynomial in $x = 2^{1/3} = \sqrt[3]{2}$, i.e. you can write it as $5 - 7x + 3x^2$. I use $x$ for simplicity (something mathematicians very often do), so everywhere you see one you can swap it for $2^{1/3}$ if you like.

We want a polynomial $p(x)$ so that $(5 - 7x + 3x^2)\cdot p(x)$ is just a rational constant term. We will get that by utilizing that $x^3 = 2$ and $x^4 = 2x$. This also means that $p(x)$ doesn't need to be more than a second degree polynomial (were it of any higher degree, we could take any higher degree term and reduce it by three degrees because of $x^3 = 2$).

So let's write down a tentative $p(x) = a + bx + cx^2$. We have $$ (5 - 7x + 3x^2)\cdot p(x) = 5a + (5b - 7a)x + (5c - 7b + 3a)x^2 + (-7c + 3b)x^3 + 3cx^4\\ = 5a - 14c + 6b + (5b -7a + 6c)x + (5c - 7b + 3a)x^2 $$ (Note that $(-7c + 3b)$ and $3c$ doubled when I moved them down three degrees. That doubling comes from $x^3 = 2$. If you are going to do this for other cube roots, e.g. $3^{1/3}$, you will have to multiply by something else, e.g. $3$.) This polynomial is supposed to be just a constant term, so we must have $$ 5b - 7a + 6c = 0 \quad \land \quad 5c - 7b + 3a = 0 $$ Since for any valid polynomial $p(x)$ we also have $k\cdot p(x)$ valid for a rational $k$, we can let $k = \frac{1}{c}$, and assume that $c = 1$. We then get $$ 7a - 5b = 6 \quad \land \quad 7b - 3a = 5 $$ with the solutions $$ b = \frac{53}{34}\quad a= \frac{67}{34} $$ so the polynomial $$ p(x) = \frac{67}{34} + \frac{53}{34}x + x^2 $$ works. This is not so pretty, though, so we multiply it by $34$ to get $$ 34\cdot p(x) = 67 + 53x + 34x^2 = 67 + 53\cdot 2^{1/3} + 34\cdot 4^{1/3} $$ So that's what we have to expand the fraction by to get a rational denominator (the denominator happens to become $177$, but that's not the important part of this exercise).

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$\frac1{A+B\sqrt[3]{n}+C\sqrt[3]{n^2}}= \frac{ \begin{vmatrix} 1&\sqrt[3]n&\sqrt[3]{n^2}\\ B&A&Cn\\ C&B&A \end{vmatrix} }{\begin{vmatrix} A&Cn&Bn\\ B&A&Cn\\ C&B&A \end{vmatrix}}= \frac{A^2-BCn+(C^2n-AB)\sqrt[3]{n}+(B^2-AC)\sqrt[3]{n^2}} {A^3+B^3n+C^3n^2-3ABCn}$

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