Solve the following system of equations: $\;$ $x + y = 9$, $\;$ $x^{\frac{1}{3}} + y^{\frac{1}{3}} = 3$
Given system of equations:
$x + y = 9$ $\;\;\; \cdots \; (1)$, $\;\;$ $x^{\frac{1}{3}} + y^{\frac{1}{3}} = 3$ $\;\;\; \cdots \; (2)$
Let $\;\;$ $x^{\frac{1}{3}} = p$ $\;\;\; \cdots \; (3)$, $\;$ $y^{\frac{1}{3}} = q$ $\;\;\; \cdots \; (4)$
Then equations $(1)$ and $(2)$ respectively become
$p^3 + q^3 = 9$ $\;\;\; \cdots \; (1a)$ $\;\;$ and $\;\;$ $p + q = 3$ $\;\;\; \cdots \; (2a)$
We have from equation $(2a)$, $\;\;$ $p = 3 - q$ $\;\;\; \cdots \; (5)$
In view of equation $(5)$, equation $(1a)$ becomes
$\left(3 - q\right)^3 + q^3 = 9$
i.e. $\;$ $27 - 27q + 9q^2 - q^3 + q^3 = 9$
i.e. $\;$ $9q^2 - 27q + 18 = 0$
i.e. $\;$ $q^2 - 3q + 2 = 0$
i.e. $\;$ $\left(q - 2\right) \left(q - 1\right) = 0$
i.e. $\;$ $q = 2$ $\;\;\;$ or $\;\;\;$ $q = 1$
When $\;$ $q = 2$, $\;$ we have from equation $(5)$, $\;$ $p = 3 - 2 = 1$
When $\;$ $q = 1$, $\;$ we have from equation $(5)$, $\;$ $p = 3 - 1 = 2$
When $\;$ $p = 1$ $\;$ and $\;$ $q = 2$,
we have from equation $(3)$, $\;$ $x^{\frac{1}{3}} = 1$ $\implies$ $x = 1$
and from equation $(4)$, $\;$ $y^{\frac{1}{3}} = 2$ $\implies$ $y = 8$
When $\;$ $p = 2$ $\;$ and $\;$ $q = 1$,
we have from equation $(3)$, $\;$ $x^{\frac{1}{3}} = 2$ $\implies$ $x = 8$
and from equation $(4)$, $\;$ $y^{\frac{1}{3}} = 1$ $\implies$ $y = 1$
$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(1, 8\right), \left(8, 1\right) \right\}$