Solve the following system of equations: $\;$ $x^3 + y^3 = 7, \;\; xy \left(x + y\right) = -2$
Given system of equations:
$x^3 + y^3 = 7$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $xy \left(x + y\right) = -2$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$\left(x + y\right)^3 - 3xy \left(x + y\right) = 7$ $\;\;\; \cdots \; (1a)$
Let $\;$ $x + y = u$ $\;\;\; \cdots \; (3a)$, $\;\;$ $xy = v$ $\;\;\; \cdots \; (3b)$
Then, in view of equations $(3a)$ and $(3b)$, equations $(2)$ and $(1a)$ respectively become
$uv = -2$ $\;\;$ i.e. $\;$ $u = \dfrac{-2}{v}$ $\;\;\; \cdots \; (4)$
and $\;\;$ $u^3 - 3uv = 7$ $\;\;\; \cdots \; (5)$
In view of equation $(4)$, equation $(5)$ can be written as
$\left(\dfrac{-2}{v}\right)^3 - 3 \times \left(\dfrac{-2}{v}\right) \times v = 7$
i.e. $\;$ $\dfrac{-8}{v^3} + 6 = 7$
i.e. $\;$ $\dfrac{-8}{v^3} = 1$ $\implies$ $v^3 = -8$ $\implies$ $v = -2$
Substituting $\;$ $v = -2$ $\;$ in equation $(4)$ gives $\;\;$ $u = \dfrac{-2}{-2} = 1$
Substituting the values of $u$ and $v$ in equations $(3a)$ and $(3b)$ gives
$x + y = 1$ $\;\;\; \cdots \; (6a)$
and $\;\;$ $xy = -2$ $\;\;\;$ i.e. $\;$ $x = \dfrac{-2}{y}$ $\;\;\; \cdots \; (6b)$
In view of equation $(6b)$, equation $(6a)$ becomes
$\dfrac{-2}{y} + y = 1$
i.e. $\;$ $y^2 - y - 2 = 0$
i.e. $\;$ $\left(y - 2\right) \left(y + 1\right) = 0$
i.e. $\;$ $y = 2$ $\;\;$ or $\;\;$ $y = -1$
Substituting the value of $y$ in equation $(6b)$ 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 $\;\;$ $\left(x, y\right) = \left\{\left(-1, 2\right), \; \left(2, -1\right) \right\}$