Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Expand the fraction by $(67 + 53\cdot 2^{1/3} + 34\cdot 4^{1/3})$. –  Arthur Feb 19 at 9:05
    
Hint: Use complex numbers. –  user2369284 Feb 19 at 9:06
    
I'd hate to think you might have to factor that denominator first. It doesn't look clean... –  Mike Feb 19 at 9:07
1  
@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. –  Arthur Feb 19 at 9:09
    
@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? –  Mike Feb 19 at 9:19

2 Answers 2

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).

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.