If ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ and ${\mathrm{bx}}^2+\mathrm{cx}+\mathrm{a}=0$ have a common root and $\mathrm{a}\ne 0.$ then $\frac{{\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3}{\mathrm{abc}}$

$\begin{array}{l}Let\alpha bethecommonrootthen\\ a{\alpha }^2+b\alpha +c=0,b{\alpha }^2+c\alpha +a=0\\ solvig\\ {\alpha }^2\alpha 1\\ \begin{array}{cccc}b& c& a& b\\ c& a& b& c\end{array}\\ \Rightarrow \frac{{\alpha }^2}{ab-c^2}=\frac{\alpha }{bc-a^2}=\frac{1}{ac-b^2}\\ \Rightarrow \frac{{\alpha }^2}{ab-c^2}=\frac{\alpha }{bc-a^2}\Rightarrow \alpha =\frac{ab-c^2}{bc-a^2}\\ and\frac{\alpha }{bc-a^2}=\frac{1}{ac-b^2}\Rightarrow \alpha =\frac{bc-a^2}{ac-b^2}\\ \therefore \frac{ab-c^2}{bc-a^2}=\frac{bc-a^2}{ac-b^2}\\ \Rightarrow a^{2}bc-a{b}^3-a{c}^3+c^2{b}^2=b^2{c}^2+a^4-2bc{a}^2\\ \Rightarrow a^4+a{b}^3+a{c}^3=3a^{2}bc\\ \Rightarrow a^3+b^3+c^3=3abc\\ \Rightarrow \frac{a^3+b^3+c^3}{abc}=3\end{array}$