simplify rational expression with roots

Question

$\frac{\sqrt[3]{25}+\sqrt[3]{5x}+\sqrt[3]{x^2} }{\sqrt[3]{x}-\sqrt[3]{5}}$

How do i simplify this and how do I get a non-square value at the bottom?

difference of cubes tells me that $a^3-b^3=(a-b)(a^2+2ab+b^2)$ and if put $\sqrt[3]{x}=a$ and $b=\sqrt[3]{5}$ the denominator vill be (a-b) and the numerator will be exactly $(a^2+2ab+b^2)$. But i can't quite seem to get it right.

any help is much appreciated!

Answer 1

Hints:

$$a^3-b^3=(a-b)(a^2+ab+b^2)\implies\frac{a^2+ab+b^2}{a-b}=\frac{a^3-b^3}{a-b}$$

Answer 2

Multiply on both numerator and denominator (a^2+b^2+a*b), where a=cube square of x and b=cube square of 5. That should be working in order to have a non-square value of the bottom.

Answer 3

We use the remarkable product:

$(A^{1/3}-B^{1/3})(A^{2/3}+(A B)^{1/3}+B^{2/3})=A-B$.

$(\sqrt[3]{A}-\sqrt[3]{B})(\sqrt[3]{A^{2}}+\sqrt[3]{A B}+\sqrt[3]{B^{2}})=A-B$

together with the principle of equivalence of fractions. "Multiplying and dividing both the numerator and the denominator of a fraction by the same quantity, one obtains a fraction equivalent to that date, that is, the value of this fraction is the same as the starting one."

The quantity in question is:

$\sqrt[3]{25}+\sqrt[3]{5x}+\sqrt[3]{x^2}$.

Therefore you have:

$\frac{\sqrt[3]{25}+\sqrt[3]{5x}+\sqrt[3]{x^2} }{\sqrt[3]{x}-\sqrt[3]{5}}=$

$=\frac{(\sqrt[3]{25}+\sqrt[3]{5x}+\sqrt[3]{x^2})^{2}}{(\sqrt[3]{x}-\sqrt[3]{5})(\sqrt[3]{25}+\sqrt[3]{5x}+\sqrt[3]{x^2})}=$

$=\frac{(\sqrt[3]{25}+\sqrt[3]{5x}+\sqrt[3]{x^2})^{2}}{x-5}$.