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,
$$a=1$$
$$b=-(\alpha^2+\alpha)$$
$$c=\alpha^3$$

Vieta's formula says:
$\alpha+\alpha^2=-\dfrac ba,\alpha\cdot\alpha^2=\dfrac ca$
$$\left(-\dfrac ba\right)^3=(\alpha+\alpha^2)^3=\alpha^3+(\alpha^3)^2+3\alpha^3\left(-\dfrac ba\right)$$
Replace $\alpha^3$ with $\dfrac ca$  and simplify
