Cancel down the fraction $\dfrac{(x-y)(\sqrt[3]{x}+\sqrt[3]{y})}{\sqrt[3]{x^2}-\sqrt[3]{y^2}}$ [closed]

Solution 1:

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:

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