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. 
 A: 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! 
A: This is a homogenous equation in third degree, representing 3 straight lines passing through (0,0)
So we have slope = $\dfrac{y}{x} =m$. We may now form cubic in $m$, having three roots $m_1, m_2, m_3$.
$$m^3+cm^2+bm+1 =0$$
From vieta theorem, we have product of roots:
$$m_1m_2m_3 = -1$$
But we also know that if two of the lines are perpendicular, then product of two of the three slopes will be $-1$. So the third slope will be $1$ and $m=1$ is a root of the cubic.
Therefore $b+c+2=0$
Edit
As Mr Rozenberg points out, we are given that $b+c = -2$, which implies that $m=1$ is a root. Since we had product of the three roots $m_1m_2m_3 = -1$ and one slope as $1$, product of other two slopes is $-1$. 
