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,
$$a=1$$
$$b=-(\alpha^2+\alpha)$$
$$c=\alpha^3$$
 A: 
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.
A: 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
A: Please, in your future exercises, try not to name different things with the same variable ($a$ in this example), because it causes a huge confusion. Also, remember than the quadratic factorization has the coefficient of $x^2$.
We have the equation : $ax^2 + bx + c = 0$
Now, let's assume it has two roots, $ρ_1 , ρ_2 $ with $ρ_2 = ρ_1^2$
Then, the polynomial will be factorized as :
$a(x-ρ_1)(x-ρ_2) = 0 \Leftrightarrow a(x-ρ_1)(x-ρ_1^2)=0 \Leftrightarrow  ax^2 - aρ_1 (ρ_1  + 1) x + aρ_1 ^3 = 0 $
You can continue on from there. 
Alternativelly, you can take on Vietta's Formulas :
$$\alpha+\alpha^2=-\dfrac ba$$
$$\alpha\cdot\alpha^2=\dfrac ca$$
and continue on from there.
A: 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.
