Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm not quite sure how to do this question. Every way that I tried doing it didn't yield an answer that is equivalent to the original question.


When I tried doing it, I ended up with $$(2x+1)^{2/3}\left(-\frac{8x+4}{2x+1}\right)$$ or with $$4(2x+1)^{1/3}$$

How can I factor it properly?

share|cite|improve this question
up vote 2 down vote accepted

Start by rewritting the equation a bit more clearly.

$$(2x+1)^{2/3} -4(2x+1)^{-1/3} = \sqrt[3]{\left ( 2x+1 \right)^2} - \frac{4}{\sqrt[3]{2x+1}}$$

Then, put everything on the same denominator.

$$\sqrt[3]{\left ( 2x+1 \right)^2} - \frac{4}{\sqrt[3]{2x+1}} = \frac{\sqrt[3]{\left ( 2x+1 \right)^2} \sqrt[3]{2x+1} - 4}{\sqrt[3]{2x+1}}$$


$$ \frac{\sqrt[3]{\left ( 2x+1 \right)^2} \sqrt[3]{2x+1} - 4}{\sqrt[3]{2x+1}} = \frac{\sqrt[3]{\left ( 2x+1 \right)^3} - 4}{\sqrt[3]{2x+1}}$$

I'll let you continue the work. Let me know in the comments if you want more info or a confirmation of your answer.

share|cite|improve this answer
More directly, factor out the smallest power of the common term, in this case $(2x+1)^{-1/3}$. – gaddy Jan 19 '14 at 4:20
I got (2x-3)/cbrt(2x+1). I think it seems right. Thanks a lot! – Alex Jan 19 '14 at 4:36
Yep, it is all right. Sorry I made a typo, $+4$ instead of $-4$. – Olivier Jan 19 '14 at 4:38

Let $\ \color{#c00}{y} = (\color{#0a0}{2x+1})^{1/\color{#c00}3}.\,$ Then $\ y^2-\dfrac{4}y\, =\, \dfrac{\color{#c00}{y^3}-4}y\, =\, \dfrac{2x-3}y\ $ by $\ \color{#c00}{y^3} = \color{#0a0}{2x+1}$

share|cite|improve this answer

First way

Multiply and divide both pieces by $(2 x + 1)^{1/3}$. You so obtain
$$\frac{2x+1 - 4}{(2 x + 1)^{1/3}} = \frac{2 x - 3}{(2 x + 1)^{1/3}}$$

Second way

Factor $(2 x + 1)^{2/3}$. You so obtain
$$ (2x+1)^{2/3}\cdot \frac{1-4}{2 x + 1} = (2 x + 1)^{2/3}\cdot \frac{2 x + 1 - 4}{2 x + 1} = \frac{2 x - 3}{(2 x + 1)^{1/3}} $$

share|cite|improve this answer
@dfeuer. Thanks for editing for me. Cheers – Claude Leibovici Jan 19 '14 at 4:45
My edit was but a slight improvement of Olivier's suggested one. – dfeuer Jan 19 '14 at 4:48
I do highly recommend using the hash-marks instead of bold for section headings so the HTML will come out with proper header elements. – dfeuer Jan 19 '14 at 4:50

As a general method you can get good results by factoring off the the binomial to the power that is leftmost on the real number line like so:

\begin{align*} (2x+1)^{\frac{2}{3}}-4(2x+1)^{-\frac{1}{3}} &= (2x+1)^{-\frac{1}{3}}\left( (2x+1)^{\frac{3}{3}} -4\right) \\ &= (2x+1)^{-\frac{1}{3}}( 2x+1 -4) \\ &= (2x+1)^{-\frac{1}{3}}( 2x-3) \\ &=\frac{2x-3}{\sqrt[3]{2x+1}}. \end{align*}

share|cite|improve this answer

Your Answer


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.