<s>[INST] 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,
$$a=1$$
$$b=-(\alpha^2+\alpha)$$
$$c=\alpha^3$$
 [/INST] 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.
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