If $a d \neq 0$ and two of the lines represented by $a x^3+3 b x^2 y$ $+3 \mathrm{cxy}^2+\mathrm{dy}^3=0$…
If $a d \neq 0$ and two of the lines represented by $a x^3+3 b x^2 y$ $+3 \mathrm{cxy}^2+\mathrm{dy}^3=0$ are perpendicular, then
$a^2+a c+b d+d^2=0$
$a^2+3 a c+3 b d+d^2=0$
$a^2-3 a c-3 b d+d^2=0$
$a^2+3 a c-3 b d+d^2=0$
Solution
$a x^3+3 b x^2 y+3 c x y^2+d y^3=0...(i)$
is a homogeneous equation of third degree is $x \& y$.
Let the slopes of the lines be $m_1, m_2, m_3$.
Then, $m_1, m_2$ and $m_3$ are the roots of
$\mathrm{dm}^3+3 \mathrm{~cm}^2+3 \mathrm{bm}+a=0 ...(ii)$
Product of roots $=m_1 m_2 m_3=-\frac{a}{d}...(iii)$
Since, the two lines represented by (i) are perpendicular angles. Let the lines with slopes $m_1, m_2$ be perpendicular
$\therefore m_1 \cdot m_2=-1$
From equation (iii), $(-1) m_3=-\frac{a}{d} \Rightarrow m_3=\frac{a}{d}...(iv)$
But $\mathrm{m}_3$ is root of equation (ii):
$\begin{aligned}
& d\left(\frac{a}{d}\right)^3+3 c\left(\frac{a}{d}\right)^2+3 b\left(\frac{a}{d}\right)+a=0 \\
& \Rightarrow a^3+3 a^2 c+3 a b d+a d^2=0
\end{aligned}$