Solve the following system of equations: $\;$ $\sqrt[3]{x} + \sqrt[3]{y} = 3$, $\;$ $\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2} = 3$
Given system of equations:
$\sqrt[3]{x} + \sqrt[3]{y} = 3$ $\;\;\; \cdots \; (1)$, $\;\;$ $\sqrt[3]{x^2} - \sqrt[3]{xy} + \sqrt[3]{y^2} = 3$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$x^{\frac{1}{3}} + y^{\frac{1}{3}} = 3$ $\;\;$ i.e. $\;\;$ $x^{\frac{1}{3}} = 3 - y^{\frac{1}{3}}$ $\;\;\; \cdots \; (3)$
Equation $(2)$ can be written as
$x^{\frac{2}{3}} - x^{\frac{1}{3}} y^{\frac{1}{3}} + y^{\frac{2}{3}} = 3$ $\;\;$ i.e. $\;\;$ $\left(x^{\frac{1}{3}}\right)^2 - x^{\frac{1}{3}} y^{\frac{1}{3}} + \left(y^{\frac{1}{3}}\right)^2 = 3$ $\;\;\; \cdots \; (4)$
In view of equation $(3)$, equation $(4)$ becomes
$\left(3 - y^{\frac{1}{3}}\right)^2 - \left(3 - y^{\frac{1}{3}}\right) y^{\frac{1}{3}} + y^{\frac{2}{3}} = 3$
i.e. $\;$ $9 + y^{\frac{2}{3}} - 6 y^{\frac{1}{3}} - 3 y^{\frac{1}{3}} + y^{\frac{2}{3}} + y^{\frac{2}{3}} = 3$
i.e. $\;$ $3 y^{\frac{2}{3}} - 9 y^{\frac{1}{3}} + 6 = 0$
i.e. $\;$ $\left(y^{\frac{1}{3}}\right)^2 - 3 y^{\frac{1}{3}} + 2 = 0$
i.e. $\;$ $\left(y^{\frac{1}{3}} - 2\right) \left(y^{\frac{1}{3}} - 1\right) = 0$
i.e. $\;$ $y^{\frac{1}{3}} = 2$ $\;\;\;$ or $\;\;\;$ $y^{\frac{1}{3}} = 1$
i.e. $\;$ $y = 8$ $\;\;\;$ or $\;\;\;$ $y = 1$
Substituting the value of $y$ in equation $(3)$ gives
when $\;$ $y = 8$, $\;$ $x^{\frac{1}{3}} = 3 - 8^{\frac{1}{3}} = 3 - 2 = 1$ $\implies$ $x = 1$
when $\;$ $y = 1$, $\;$ $x^{\frac{1}{3}} = 3 - 1^{\frac{1}{3}} = 3 - 1 = 2$ $\implies$ $x = 8$
$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(1, 8\right), \; \left(8, 1\right) \right\}$