Prove that the equation $\;$ $y^3 - x^3 + 3xy \left(y - x\right) = 0$ $\;$ represents three straight lines equally inclined to one another.
Given equation is: $\;$ $y^3 - x^3 + 3xy \left(y - x\right) = 0$ $\;\;\; \cdots \; (1a)$
Equation $(1a)$ can be written as
$\left(y - x\right) \left(y^2 + xy + x^2\right) + 3xy \left(y - x\right) = 0$
i.e. $\;$ $\left(y - x\right) \left(y^2 + xy + x^2 + 3xy\right) = 0$
i.e. $\;$ $\left(y - x\right) \left(x^2 + 4xy + y^2\right) = 0$ $\;\;\; \cdots \; (1b)$
Equation $(1b)$ represents the pair of lines
$y - x = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $x^2 + 4xy + y^2 = 0$ $\;\;\; \cdots \; (2b)$
Equation $(2a)$ can be written as $\;\;$ $y = x$
Comparing with the standard equation $\;$ $y = mx + c$ $\;$ (c is the intercept on the Y axis) $\;$ gives
slope of equation $(2a)$ $= m = 1$ $\;\;\; \cdots \; (3a)$
Comparing equation $(2b)$ with the standard equation $\;\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives
$a = 1, \; h = 2, \; b = 1$ $\;\;\; \cdots \; (3b)$
Let $\theta$ be the angle between the pair of straight lines given by equation $(2b)$.
Then,
$\begin{aligned}
\tan \theta & = \dfrac{2 \sqrt{h^2 - ab}}{a + b} \\\\
& = \dfrac{2 \sqrt{2^2 - 1 \times 1}}{1 + 1} \\\\
& = \dfrac{2 \sqrt{3}}{2} = \sqrt{3}
\end{aligned}$
$\implies$ $\theta = \tan^{-1} \left(\sqrt{3}\right) = 60^\circ$ $\;\;\; \cdots \; (4a)$
Let $m_1$ and $m_2$ be the slopes of the lines given by equation $(2b)$.
Then, $\;$ $m_1 + m_2 = \dfrac{-2h}{b} = \dfrac{-2 \times 2}{1} = -4$ $\;\;\; \cdots \; (5a)$
and $\;$ $m_1 - m_2 = \dfrac{2 \sqrt{h^2 - ab}}{\left|b\right|} = \dfrac{2 \sqrt{2^2 - 1 \times 1}}{\left|1\right|} = 2 \sqrt{3}$ $\;\;\; \cdots \; (5b)$
From equations $(5a)$ and $(5b)$,
$2 m_1 = -4 + 2 \sqrt{3}$ $\implies$ $m_1 = \sqrt{3} - 2$ $\;\;\; \cdots \; (6a)$
and $\;$ $m_2 = -4 - m_1 = -4 - \left(\sqrt{3} - 2\right) = -\sqrt{3} - 2$ $\;\;\; \cdots \; (6b)$
Let $\phi$ be the angle between the lines with slopes $m$ and $m_1$.
Then,
$\begin{aligned}
\tan \phi & = \left|\dfrac{m - m_1}{1 + m m_1}\right| \\\\
& = \left|\dfrac{1 - \left(\sqrt{3} - 2\right)}{1 + 1 \times \left(\sqrt{3} - 2\right)}\right| \;\; \left[\text{by equations (3a) and (6a)}\right] \\\\
& = \left|\dfrac{3 - \sqrt{3}}{1 - 2 + \sqrt{3}}\right| \\\\
& = \dfrac{3 - \sqrt{3}}{\sqrt{3} - 1} \\\\
& = \dfrac{\left(3 - \sqrt{3}\right) \left(\sqrt{3} + 1\right)}{\left(\sqrt{3} - 1\right) \left(\sqrt{3} + 1\right)} \\\\
& = \dfrac{3 \sqrt{3} + 3 - 3 - \sqrt{3}}{3 - 1} = \dfrac{2 \sqrt{3}}{2} = \sqrt{3}
\end{aligned}$
$\implies$ $\phi = \tan^{-1} \left(\sqrt{3}\right) = 60^\circ$ $\;\;\; \cdots \; (4b)$
Let $\gamma$ be the angle between the lines with slopes $m$ and $m_2$.
Then,
$\begin{aligned}
\tan \gamma & = \left|\dfrac{m - m_2}{1 + m m_2}\right| \\\\
& = \left|\dfrac{1 - \left(-\sqrt{3} - 2\right)}{1 + 1 \times \left(-\sqrt{3} - 2\right)}\right| \;\; \left[\text{by equations (3a) and (6b)}\right] \\\\
& = \left|\dfrac{3 + \sqrt{3}}{1 - 2 - \sqrt{3}}\right| \\\\
& = \dfrac{3 + \sqrt{3}}{\sqrt{3} + 1} \\\\
& = \dfrac{\left(3 + \sqrt{3}\right) \left(\sqrt{3} - 1\right)}{\left(\sqrt{3} - 1\right) \left(\sqrt{3} + 1\right)} \\\\
& = \dfrac{3 \sqrt{3} - 3 + 3 - \sqrt{3}}{3 - 1} = \dfrac{2 \sqrt{3}}{2} = \sqrt{3}
\end{aligned}$
$\implies$ $\gamma = \tan^{-1} \left(\sqrt{3}\right) = 60^\circ$ $\;\;\; \cdots \; (4c)$
$\therefore \;$ We have from equations $(4a)$, $(4b)$ and $(4c)$, $\;$ $\theta = \phi = \gamma$
$\implies$ The three straight lines given by equation $(1a)$ are equally inclined to one another.
Hence proved.