What are the equations of the three straight lines represented by $x^3+bx^2y+cxy^2+y^3=0$ when $b+c=-2$? I am given that $$x^3 + bx^2y + cxy^2 + y^3 = 0$$ represents three straight lines if $b + c = -2$.
Is there a way to find the equations of the three lines separately?
I tried factorizing the equation but wasn't able to get anywhere. Is there a way to get the separate equations using partial differentiation?
 A: If $b+c=-2$, you can write $b=-k$ and $c=k-2$, so
$$\begin{align}x^3+bx^2y+cxy^2+y^3&=x^3-kx^2y+(k-2)xy^2+y^3\\&=x^3-kx^2y+kxy^2-2xy^2+y^3\\&=(x^3-xy^2)-(kx^2y-kxy^2)-(xy^2-y^3)\\&=(x-y)(x(x+y)-kxy-y^2)\\&=(x-y)(x^2-(k-1)xy-y^2)\end{align}$$
You can factor the latter part by completing the square but the computation is a bit ugly.
The three lines will be:
$$x=y\\ x=y\left(\frac{k-1}2+\sqrt{1+\left(\frac{k-1}2\right)^2}\right)\\ x=y\left(\frac{k-1}2-\sqrt{1+\left(\frac{k-1}2\right)^2}\right)$$
where $k$ is just $-b$, a constant, so the entire expression in the brackets is a constant, the reciprocal of their slopes.
A: Observe that all three straight lines pass through the origin $(0, 0)$
Let the three straight lines are
$y = m_1x$, $y = m_2x$ and $y = m_3x$
Then
$$ (y -m_1x)(y -m_2x)(y -m_3x) \equiv x^3 + bx^2y + cxy^2 + y^3$$
Comparing the coefficients, we have
$m_1 m_2 m_3 = -1$
$m_1 m_2 + m_1 m_3 + m_2 m_3 = b$
$m_1 + m_2 + m_3 = -c$
These are the three roots of the equation
$m^3 + cm^2 + bm +1 = 0$
If $b + c = -2$, then $m=1$ must be a root of the above equation. The other quadratic factor is $m^2 - bm - m - 1$ and has the roots $\frac{1}{2} (1 + b \pm \sqrt{b^2 + 2 b + 5})$
Finally the three straight lines are
$y = x$
$y = \frac{x}{2} (1 + b + \sqrt{b^2 + 2 b + 5})$
and
$y = \frac{x}{2} (1 + b - \sqrt{b^2 + 2 b + 5})$
