If $x^5-bx^3+cx^2+dx-e$ can be expressed as the product of a perfect square and a perfect cube then prove that $$\frac{12b}{5}=\frac{9d}{b}=\frac{5e}{c}=\frac{d^2}{c^2}$$
My attempt is as follows: $$E=x^5-bx^3+cx^2+dx-e$$ $$E=(x-\alpha)^2(x-\beta)^3$$ $$E=x^5-\left(2\alpha+3\beta\right)x^4+\left(3{\alpha}^2+6\alpha\beta+{\beta}^2\right)x^3-\left({\alpha}^3+6{\alpha}^2\beta+3\alpha{\beta}^2\right)x^2+\left(2{\alpha}^3\beta+3{\alpha}^2{\beta^2}\right)x-{\alpha}^2{\beta}^3$$ $$x^5-bx^3+cx^2+dx-e=x^5-\left(2\alpha+3\beta\right)x^4+\left(3{\alpha}^2+6\alpha\beta+{\beta}^2\right)x^3-\left({\alpha}^3+6{\alpha}^2\beta+3\alpha{\beta}^2\right)x^2+\left(2{\alpha}^3\beta+3{\alpha}^2{\beta^2}\right)x-{\alpha}^2{\beta}^3$$
So we obtain five equations
$$\left(2\alpha+3\beta\right)=0$$ \begin{equation} \frac{\alpha}{\beta}=\frac{-3}{2}\tag{1} \end{equation}
$$3{\alpha}^2+6\alpha\beta+{\beta}^2=-b$$ $$\text {Dividing by ${\beta}^2$}$$ $$3\frac{\alpha^2}{\beta^2}+6\frac{\alpha}{\beta}+1=\frac{-b}{{\beta}^2}$$ \begin{equation} \frac{b}{{\beta}^2}=\frac{5}{4}\tag{2} \end{equation}
$${\alpha}^3+6{\alpha}^2\beta+3\alpha{\beta}^2=-c$$ $$\text {Dividing by ${\beta}^3$}$$ $$\frac{\alpha^3}{\beta^3}+6\frac{\alpha^2}{\beta^2}+3\frac{\alpha}{\beta}=\frac{-c}{{\beta}^3}$$ \begin{equation} \frac{c}{{\beta}^3}=\frac{-45}{8}\tag{3} \end{equation}
$$2{\alpha}^3\beta+3{\alpha}^2{\beta}^2=d$$ $$\text {Dividing by ${\beta}^4$}$$ $$2\frac{\alpha^3}{\beta^3}+3\frac{\alpha^2}{\beta^2}=\frac{d}{{\beta}^4}$$ \begin{equation} \frac{d}{{\beta}^4}=0\tag{4} \end{equation}
So $d$ is coming as $0$ and this is where I am stuck because for $$\frac{12b}{5}=\frac{9d}{b}=\frac{5e}{c}=\frac{d^2}{c^2}$$ $d$ should be non-zero because $\dfrac{12b}{5}=3\beta^2$ is non zero.
Please help me.
Since $$\alpha=-\frac{3}{2}\beta,$$ we obtain: $$E=\left(x+\frac{3}{2}\beta\right)^2(x-\beta)^3=x^5-\frac{15}{4}\beta^2x^3+\frac{5}{4}\beta^3x^2+\frac{15}{4}\beta^4x-\frac{9}{4}\beta^5,$$ which gives $$b=\frac{15}{4}\beta^2,$$ $$c=\frac{5}{4}\beta^3,$$ $$d=\frac{15}{4}\beta^4$$ and $$e=\frac{9}{4}\beta^5.$$ Can you end it now?
For example, we need to prove that: $$12bc^2=5d^2$$ or $$12\cdot\frac{15}{4}\beta^2\cdot\left(\frac{5}{4}\beta^3\right)^2=5\cdot\left(\frac{15}{4}\beta^4\right)^2.$$