Rewrite 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 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!

Hint:$a^3-b^3=(a-b)(a^2+ab+b^2)$
Let $a=x^{1/3},b=7^{1/3}$
$x-7=a^3-b^3=(a-b)(a^2+ab+b^2)=(x^{1/3}-7^{1/3})(7^{2/3}+(7x)^{1/3}+x^{2/3})$
$\displaystyle \frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(x^{1/3}-7^{1/3})}\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})}{(7^{2/3}+(7x)^{1/3}+x^{2/3})}=\frac{(7^{2/3}+(7x)^{1/3}+x^{2/3})^2}{(x-7)}$
