Doubt on whether my solution is correct

Question

The question was to prove that the equation $y^3-x^3+3xy(y-x)=0$ represents three lines equally inclined to one another.What i did was i factored out $(y-x)$ and the equation became $$(y-x)(y^2+4xy+x^2)$$. Now, since the other part is a homogeneous equation of second degree, so i can safely say that it represents two straight lines. Now i guessed that if the lines are equally inclined then the bisector of the lines given by $y^2+4xy+x^2$ will be the line $y-x=0$. So, going forward to prove it i tried to find the bisectors by using the formula $(x^2-y^2)/(a-b)=xy/h$, where a and b are the coefficients of $x^2$ and $y^2$ and h is the coeffient of the term $2xy$. The equations of the bisectors were $y-x$ and $y+x$. which matched the factored out term in the original equation.So, my question is that , is this enough to complete the proof?

Answer 1

1) The formula used should be $hx^2 –(a – b)xy – h^2 = 0$ instead. This is because one has to avoid the $a – b$ term being the zero denominator.

2) It turns out one of the angle bisectors is $L_1 : y = x$. This means both $L_2$ & $L_3$ are equally inclined to $L_1$.

3) We still need to prove that $L_2$ or $L_3$ are also equal inclined to $L_1$.

This can be done by:-

(a) Finding the angle between $L_2$ and $L_3$ via the formula $\tan \theta = \dfrac {2\sqrt {h^2 - ab}}{a + b}$. (It turns out that $\theta = 60^0$); and

(b) Further factorizing $y^2+4xy+x^2$ to $[(y + 2x) + \sqrt 3 x][ (y + 2x) - \sqrt 3x]$. Then, find the angle between $L_1$ and $L_2$.