Prove that two of the straight lines represented by $x^3+bx^2y+cxy^2+y^3 = 0$ will be at right angles if $b+c$=-2. 
Prove that two of the straight lines represented by $x^3+bx^2y+cxy^2+y^3 = 0$ will be at right angles if $b+c=-2$.

I divided whole equation by $y^3$ to get a cubic in $\frac{x}{y}$. I guess product of two roots of this equation should be $-1$ but I am unable to derive a condition for that. 

Let $y=mx$.
Thus, me need
$$m^3+bm^2+cm+1=0$$ with $m_1m_2=-1$ and since $m_1m_2m_3=-1$, 
we obtain $m_3=1$ and $$1+b+c+1=0$$ or
$$b+c=-2.$$
Id est, we proved a bit of more:

Two of the straight lines represented by  $$x^3+bx^2y+cxy^2+y^3=0$$  will be at right angles iff $b+c=-2$. 

Because if $b+c=-2$ we see that one of roots of the following equation
$$m^3+bm^2+cm+1=0$$ is $m_1=1$, which gives $m_2m_3=-1$ and we are done! 
