If $a>b>0$ and $a^3 +b^3 +27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $P,Q(P
If $a>b>0$ and $a^3 +b^3 +27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $P,Q(P<Q)$. Find the value of $4Q-aP$?
How will I begin with the solution just a hint would be enough.

$$a^3 +b^3 +(-9)^3-3ab(-9)=(a+b-9)(a^2+b^2-ab+9a+9b+81)=0$$
therefore
\begin{cases}
a+b-9=0\\
\qquad\operatorname{or}\\
a=b=-9
\end{cases}
since $a>b>0$ thus
$$a+b-9=0$$
Set $f(x)=ax^2+bx-9$. We have $f(1)=a+b-9=0$, thus 
$Q=1$ and $P=\frac{-9}{a}$
finally
$$4Q-aP=4+9=13$$
