Solve the following system of equations: $\;$ $\sqrt[3]{x} - \sqrt[3]{y} = 2$, $\;$ $xy = 27$
Given system of equations:
$\sqrt[3]{x} - \sqrt[3]{y} = 2$ $\;\;\; \cdots \; (1)$, $\;\;$ $xy = 27$ $\;\;\; \cdots \; (2)$
We have from equation $(2)$, $\;\;\;$ $x = \dfrac{27}{y}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(1)$ becomes
$\sqrt[3]{\dfrac{27}{y}} - \sqrt[3]{y} = 2$
i.e. $\;$ $\dfrac{3}{\sqrt[3]{y}} - \sqrt[3]{y} = 2$ $\;\;\; \cdots \; (4)$
Let $\;\;\;$ $\sqrt[3]{y} = p$ $\;\;\; \cdots \; (5)$
Then equation $(4)$ becomes
$\dfrac{3}{p} - p = 2$
i.e. $\;$ $p^2 + 2p - 3 = 0$
i.e. $\;$ $\left(p + 3\right) \left(p - 1\right) = 0$
i.e. $\;$ $p = -3$ $\;\;\;$ or $\;\;\;$ $p = 1$
Case 1:
When $\;$ $p = -3$, $\;$ we have from equation $(5)$
$\sqrt[3]{y} = -3$ $\implies$ $y = -27$
Substituting $\;$ $y = -27$ $\;$ in equation $(2)$ gives
$-27x = 27$ $\implies$ $x = -1$
Case 2:
When $\;$ $p = 1$, $\;$ we have from equation $(5)$
$\sqrt[3]{y} = 1$ $\implies$ $y = 1$
Substituting $\;$ $y = 1$ $\;$ in equation $(2)$ gives $\;\;\;$ $x = 27$
$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(27, 1\right), \left(-1, -27\right) \right\}$