Solve the following system of equations: $\;$ $x^3 + y^3 = 7, \;\; x^3 y^3 = -8$
Given system of equations:
$x^3 + y^3 = 7$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x^3 y^3 = -8$ $\;\;\; \cdots \; (2)$
Equation $(2)$ can be written as
$\left(xy\right)^3 = \left(-2\right)^3$
i.e. $x \; y = -2$
i.e. $\;$ $x = \dfrac{-2}{y}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(1)$ becomes
$\left(\dfrac{-2}{y}\right)^3 + y^3 = 7$
i.e. $\;$ $\dfrac{-8}{y^3} + y^3 = 7$
i.e. $\;$ $y^6 - 7y^3 - 8 = 0$ $\;\;\; \cdots \; (4)$
Let $\;\;$ $y^3 = p$ $\;\;\; \cdots \; (5)$
The equation $(4)$ becomes
$p^2 - 7p - 8 = 0$
i.e. $\;$ $\left(p - 8\right) \left(p + 1\right) = 0$
i.e. $\;$ $p = 8$ $\;\;$ or $\;\;$ $p = -1$
Substituting the value of $p$ in equation $(5)$ gives
when $\;$ $p = 8$, $\;$ $y^3 = 8$ $\;\;$ i.e. $\;$ $y = 2$
and when $\;$ $p = -1$, $\;$ $y^3 = -1$ $\;\;$ i.e. $\;$ $y = -1$
Substituting the value of $y$ in equation $(3)$ gives
when $\;\;$ $y = 2$, $\;\;$ $x = \dfrac{-2}{2} = -1$
and when $\;\;$ $y = -1$, $\;\;$ $x = \dfrac{-2}{-1} = 2$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(-1, 2\right), \; \left(2, -1\right) \right\}$