Finding a relationship between coefficients to solve a cubic polynomial From E.J Barbeau's Polynomials, the question states:
Find the relationship between $p$ and $q$ in order that the equation $x^3 + px + q$ may be put into the form $x^4 = (x^2 + ax + b)^2$. Hence, solve the equation $8x^3 - 36x + 27 = 0$.
I have no clue where to start on this. Could anyone assist me with this problem? The answer ($p^3 + 8q^2 = 0$) seems rather random and I don't understand how one could have obtained such a relationship.
 A: You want $x^3+px+q=0$ to be equivalent to $x^4=(x^2+ax+b)^2$. Notice that:
$$x^4=(x^2+ax+b)^2 \iff (x^2)^2-(x^2+ax+b)^2=0$$
$$\iff (x^2-(x^2+ax+b))(x^2+(x^2+ax+b))=0 \iff (ax+b)(2x^2+ax+b)=0$$
$$\iff 2ax^3+a^2x^2+abx+2bx^2+abx+b^2=0$$
$$\iff 2ax^3+(a^2+2b)x^2+2abx+b^2=0 \iff x^3+\frac{a^2+2b}{2a}x^2+bx+\frac{b^2}{2a}=0$$
In order for this to be equivalent to $x^3+px+q=0$, we must have equal coefficients:
$$\begin{cases}
\frac{a^2+2b}{2a}=0 \implies a^2+2b=0\\ 
p=b \implies b=p\\ 
q=\frac{b^2}{2a} \implies a=\frac{b^2}{2q} \implies a=\frac{p^2}{2q}
\end{cases}
$$
Plugging in the last two equations into the first one yields:
$$(\frac{p^2}{2q})^2+2p=0 \implies \frac{p^4}{4q^2}+2p=0 \implies p^4+8pq^2=0 \implies\fbox{$p^3+8q^2=0$}$$
Now for solving $8x^3-36x+27=0 \iff x^3-\frac{9}{2}x+\frac{27}{8}$, we can see that our condition is satisfied, hence we can turn the equation into one of the form $x^4=(x^2+ax+b)^2$. The above equations enable us to quickly find $a$ and $b$:
$$a=\frac{p^2}{2q}=\frac{\frac{9^2}{2^2}}{2\cdot\frac{27}{8}}=3$$
$$b=p=-\frac{9}{2}$$
So our equation becomes:
$$x^4=(x^2+3x-\frac{9}{2})^2 \implies x^2=\pm (x^2+3x-\frac{9}{2})$$
For the $+$ case we get:
$$x^2=x^2+3x-\frac{9}{2} \implies 3x=\frac{9}{2} \implies x_1=\frac{3}{2}$$
For the $-$ case we get:
$$x^2=-x^2-3x+\frac{9}{2} \implies 2x^2+3x-\frac{9}{2}=0 \implies x_{2,3}=\frac{-3\pm 3\sqrt{5}}{4}$$
To conclude, the solutions for $8x^3-36x+27=0$ are:
$$\fbox{$x_1=\frac{3}{2} \space \text{and} \space x_{2,3}=\frac{-3\pm 3\sqrt{5}}{4}$}$$
