If $a>b>0$ and $a^3+b^3+27ab=729$ then $ax^2+bx-9=0$ has roots $\alpha,\beta,(\alpha<\beta)$. Find the value of $4\beta -a\alpha$.

Question

If $a>b>0$ and $a^3+b^3+27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $\alpha,\beta,(\alpha<\beta)$. Find the value of $4\beta -a\alpha$.

By looking at the equation I figured out $a+b=9$. Hence one root of of the equation is 1. But I don't know how to proceed further. It would be great if I could get some help with this question.

Answer 1

Hint ( I haven't completely verified) $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ If this equal zero then $$(a+b+c)=0$$ or $$(a^2+b^2+c^2-ab-bc-ca)=0$$