<s>[INST] If one root of the equation $ax^2+bx+c=0$ be the square of the other. If $a \neq 0$ and if one root of the equation $ax^2+bx+c=0$ is the square of the other, prove that: $$b^3+a^2c+ac^2=3abc.$$
My Attempt:
Given:
$$ax^2 + bx + c=0$$
Let $\alpha $ and $\beta $ be the roots of the equation.
$$\alpha + \beta = \dfrac {-b}{a}$$
$$\alpha . \beta = \dfrac {c}{a}$$
According to the question:
$$\alpha = \beta^2 $$
 [/INST] Alternatively: 
$$\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2=\frac{-b+\sqrt{b^2-4ac}}{2a} \Rightarrow$$
$$b^2-2ac+ab=(a-b)\sqrt{b^2-4ac} \Rightarrow$$
Squaring and then dividing both sides by $4a$ will prove the claim.
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