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


Let one root be $\alpha $ then the other root will be $\alpha^2$. Then,
  $$(x-\alpha)(x-\alpha^2)=0$$

Note that you need a factorization of the form $\color{red}{a}(x-\alpha)(x-\alpha^2)=0$ to be able to compare with $ax^2+bx+c=0$. Alternatively, you can divide by $a$ and continue with a quadratic where $a=1$.
Other than that, the approach is fine. Insert the expressions you find for $a$, $b$ and $c$ into the three options. But carefully looking at the powers, you can already see 1. (because of $c^3$, leaving a term in $\alpha^9$) and 2. (because of $b^2c$, leaving a term in $\alpha^7$) won't simplify to $0$. That leaves calculating expression 3 and concluding it's either 3. or 4.

Alternatively, still calling the roots $\alpha$ and $\alpha^2$, we know that the sum and product of the roots of $ax^2+bx+c=0$ are equal to $-\tfrac{b}{a}$ and $\tfrac{c}{a}$ respectively, so:
$$-\frac{b}{a}=\alpha+\alpha^2 \quad \mbox{and} \quad \frac{c}{a}=\alpha^3$$
And continue from there.
