If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true? If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true?
$1$. $a^3+b^3+c^3-3abc=0$
$2$. $a^3+b^3+bc^2=3abc$
$3$. $b^3+a^2c+ac^2=3abc$
$4$. none.
My Attempt: 
Let one root be $\alpha $ then the other root will be $\alpha^2$. Then,
$$(x-\alpha)(x-\alpha^2)=0$$
$$x^2-x(\alpha^2+\alpha)+\alpha^3=0$$ Comparing with $ax^2+bx+c=0$ we get,
$$a=1$$
$$b=-(\alpha^2+\alpha)$$
$$c=\alpha^3$$

Product of roots $= \alpha^3 = \dfrac{c}{a}$
Now $a\alpha^2+b\alpha+c = 0 \Rightarrow a^3 \alpha^6+b^3 \alpha^3+c^3 = 3abc \alpha^3$
Substitute for $\alpha^3$ in the above and after simplification we obtain $b^3+a^2c+ac^2=3abc$
In the above we assume that $\alpha \ne 0$, but its easy to see that the above relation holds even when $\alpha = 0$ and hence all cases are covered.
