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I am struggling with rewriting the following so that the denominator does not have any root expressions:

$$\frac{\sqrt[3]{49} +\sqrt[3]{7x} + \sqrt[3]{x^2}}{\sqrt[3]{x} -\sqrt[3]{7}}$$

I guess I should start with the denominator and try to get rid of the cube root expressions. But I cannot really get how one would do that easily. Is there another way to solve this problem?

Thank you kindly for your help!

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    $\begingroup$ Remember $x^3-y^3=(x-y)(x^2+xy+y^2)$. $\endgroup$ – Maesumi Feb 24 '13 at 15:47
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    $\begingroup$ @Maesumi Post it as an answer? $\endgroup$ – Git Gud Feb 24 '13 at 15:50
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Hint:$a^3-b^3=(a-b)(a^2+ab+b^2)$

Let $a=x^{1/3},b=7^{1/3}$

$x-7=a^3-b^3=(a-b)(a^2+ab+b^2)=(x^{1/3}-7^{1/3})(7^{2/3}+(7x)^{1/3}+x^{2/3})$

$\displaystyle \frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(7^{2/3}+(7x)^{1/3}+x^{2/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})^2}{(x-7)}$

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  • $\begingroup$ Thank you for your answer! I would very much appreciate if you would write the full answer. I want to study in what order you would do the arithmetic. $\endgroup$ – Lukas Arvidsson Feb 24 '13 at 15:58
  • $\begingroup$ Now is it okay?????? @LukasArvidsson $\endgroup$ – Abhra Abir Kundu Feb 24 '13 at 16:04
  • $\begingroup$ You are welcome @LukasArvidsson $\endgroup$ – Abhra Abir Kundu Feb 24 '13 at 16:06
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Since $u^3-v^3=(u-v)(u^2+uv+v^2)$, then $$\frac{v^2+uv+u^2}{u-v}=\frac{u^2+uv+v^2}{u-v}=\frac{\left(u^2+uv+v^2\right)^2}{u^3-v^3}.$$ What are $u$ and $v$ in your particular case?

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