If $ax^2+xy+y^2+bx+cy+d = 0$ represents a pair of straight lines with slopes m and m^2 , then number of values of a are?

Question

This is a question I found: If $ax^2+xy+y^2+bx+cy+d = 0$ represents a pair of straight lines with slopes $m$ and $m^2$, then number of values of $a$ are?

The answer given is 1.

How do I approach this problem? I tried solving this by equating coefficients of $(mx-y+c_1)(m^2x-y+c_2)+k = 0$ with the given equation and got $m^3= m\cdot m^2 = a$ and $m+m^2=-1$, but I don't know what to do after this. What should I do?

Answer 1

$$m^2+m+1=0$$

$$m^3-1=(m-1)(m^2+m+1)=?$$

Answer 2

The equation $m^2+m+1=0$ has no real solutions.

Slopes are real numbers.

Hence there are no values of $a$ which satisfy the requirements.

Thus, the answer of $1$ is not correct.

Probably a typo in the problem statement, or else in the answer key.