<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] $$\alpha + \beta = \dfrac {-b}{a}$$
$$\alpha . \beta = \dfrac {c}{a}$$
Substitute
$$\alpha = \beta^2$$
into previous equations,
$$a(\beta^2+\beta) = a\beta(\beta+1) = - b\tag{1}$$
$$a\beta^3 = c\tag{2}$$ 
Cube equation $(1)$,
$$a^3 \beta^3 (\beta^3 + 3 \beta^2 + 3 \beta + 1) = -b^3$$
Using equation $(2)$,
$$a^2 c (\beta^3 + 3 \beta^2 + 3 \beta + 1) = -b^3$$
$$a c (a\beta^3 + 3a\beta( \beta + 1) + a) = -b^3$$
Using equation $(1)$ and $(2)$,
$$a c (c - 3b + a) = -b^3$$
$$ac^2-3abc+a^2c=-b^3$$
$$b^3+ac^2+a^2c = 3abc$$
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