Cancel down the following fraction $$\dfrac{(x-y)(\sqrt[3]{x}+\sqrt[3]{y})}{\sqrt[3]{x^2}-\sqrt[3]{y^2}}$$
I am not sure if cancel down means rationalize, so are we supposed to rationalize the given fraction or just cancel it down? How can we do just cancel it down?
On
I think fractional exponents are more intuitive than radicals.
$\frac {(x-y)(x^\frac 13 + y^\frac 13)}{x^\frac 23 - y^\frac23}$
We have a difference of squares in the denominator
$\frac {(x-y)(x^\frac 13 + y^\frac 13)}{(x^\frac 13 - y^\frac13)(x^\frac 13 + y^\frac13)}$
Giving us something to cancel.
$\frac {(x-y)}{(x^\frac 13 - y^\frac13)}$
We don't like radicals (or fractional roots) in the denominator but we can clean that up if we multiply top and bottom by the correct factor.
$\frac {(x-y)}{(x^\frac 13 - y^\frac13)}\frac{(x^\frac23 + x^\frac 13y^\frac 13 + y^\frac 23)}{(x^\frac23 + x^\frac 13y^\frac 13 + y^\frac 23)} = \frac {(x-y)(x^\frac23 + x^\frac 13y^\frac 13 + y^\frac 23)}{x-y} $
$x^\frac23 + x^\frac 13y^\frac 13 + y^\frac 23$ is as simple as I can get it.
Using the fact that $x^{\frac{2}{3}}=\left(x^{\frac{1}{3}}\right)^2$ , so as for $y$, we have: $$x^{\frac{2}{3}}-y^{\frac{2}{3}}=\left(x^{\frac{1}{3}}-y^{\frac{1}{3}}\right)\cdot\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)$$ Plugging in, we have: $$\dfrac{(x-y)(\sqrt[3]{x}+\sqrt[3]{y})}{\left(x^{\frac{1}{3}}-y^{\frac{1}{3}}\right)\cdot\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)}=\frac{x-y}{x^{\frac{1}{3}}-y^{\frac{1}{3}}}$$ Then, you can observe: $$x-y=\left(x^{\frac{1}{3}}-y^{\frac{1}{3}}\right)\cdot\left(x^{\frac{2}{3}}+(x\cdot y)^{\frac{1}{3}}+y^{\frac{2}{3}}\right)$$ So, finally:
$$\dfrac{(x-y)(\sqrt[3]{x}+\sqrt[3]{y})}{\sqrt[3]{x^2}-\sqrt[3]{y^2}}=\frac{\left(x^{\frac{1}{3}}-y^{\frac{1}{3}}\right)\cdot\left(x^{\frac{2}{3}}+(x\cdot y)^{\frac{1}{3}}+y^{\frac{2}{3}}\right)}{x^{\frac{1}{3}}-y^{\frac{1}{3}}}=\left(x^{\frac{2}{3}}+(x\cdot y)^{\frac{1}{3}}+y^{\frac{2}{3}}\right)$$