Rationalizing the denominator 3 It is a very difficult question. How can we Rationalizing the denominator?
$$\frac{2^{1/2}}{5+3*(4^{1/3})-7*(2^{1/3})}$$
 A: 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).
A: $\frac1{A+B\sqrt[3]{n}+C\sqrt[3]{n^2}}=
\frac{ \begin{vmatrix}
1&\sqrt[3]n&\sqrt[3]{n^2}\\ 
B&A&Cn\\ 
C&B&A
\end{vmatrix} }{\begin{vmatrix}
A&Cn&Bn\\ 
B&A&Cn\\ 
C&B&A
\end{vmatrix}}=
\frac{A^2-BCn+(C^2n-AB)\sqrt[3]{n}+(B^2-AC)\sqrt[3]{n^2}}
{A^3+B^3n+C^3n^2-3ABCn}$  
