Reminds of Newton?

$a+b+c=1$ ..... $(1)$
$a^2+b^2+c^2=4$ ...... $(2)$
$a^3+b^3+c^3=9$ ...... $(3)$

Squring both sides to $(1)$ .
$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)=1$

$\Rightarrow (ab+bc+ca)=\boxed{\dfrac{-3}{2}}$

Adding $(-3abc)$ both sides to $(3)$ .

$\Rightarrow a^3+b^3+c^3-3abc=9-3abc$

$\Rightarrow (abc)=\boxed{\dfrac{7}{6}}$ ..... $(4)$

$\Rightarrow \dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{(ab)^3+(bc)^3+(ca)^3}{(abc)^3}$

Now, finding $(ab)^3+(bc)^3+(ca)^3$ .

Let $ab=x$ , $bc=y$ and $ca=z$ .

$\Rightarrow x^3+y^3+z^3=f$

Adding $(-3xyz)$ both sides.

$\Rightarrow (x+y+z)(x^2+y^2+z^2-(xy+yz+zx)=f-3xyz$

$\Rightarrow (ab+bc+ca)[(ab+bc+ca)^2-3abc(a+b+c)]=f-3(abc)^2$

$\Rightarrow (\dfrac{-3}{2})[\dfrac{49}{36}-\dfrac{21}{6}]=f-\dfrac{49}{12}$

$\Rightarrow (ab)^3+(bc)^3+(ca)^3=\boxed{\dfrac{143}{24}}$ .... $(5)$

$\Rightarrow \dfrac{(ab)^3+(bc)^3+(ca)^3}{(abc)^3}$

Now putting values from $(4)$ and $(5)$ .

$\Rightarrow \dfrac{143}{24}×\dfrac{(7)^3}{(6)^3}$

$\Rightarrow \dfrac{30888}{8232}=\dfrac{1287}{343}$

$\Rightarrow \dfrac{1287}{343}=\dfrac{p}{q}$

$\Rightarrow p+q=1287+343=\boxed{1630}$ .

Let $a,b,c$ be the roots of the cubic equation with symmetric sums $S_{1}, S_{2}, S_{3}$ .
Let $P_{n} = a^{n} + b^{n} + c^{n}$
According to Newton's Identities,
$P_{1} = S_{1}$
$P_{2} = S_{1}P_{1} - 2S_{2}$
$P_{3} = S_{1}P_{2} - S_{2}P_{1} + 3S_{3}$
Solving these equations for $S_{1} , S_{2}, S_{3}$ yields
$S_{1} = 1, S_{2} = \dfrac{-3}{2}, S_{3} = \dfrac{7}{6}$
Therefore, the cubic equation with roots $a,b,c$ as roots is given by
$x^{3} - S_{1}x^{2} + S_{2}x - S_{3} = 0$
Substituting the values of $S_{1}, S_{2} , S_{3}$

$\therefore 6x^{3} - 6x^{3} - 9x - 7 = 0$
Thus, the equation whose roots are reciprocal of the given equation are given by substituting $x = \dfrac{1}{y}$
New equation : $7x^{3} + 9x^{2} + 6x - 6 = 0$
Let the symmetric sums of this new equation be denoted by $e_{1}, e_{2}, e_{3}$
By Vieta's formula,
$e_{1} = \dfrac{-9}{7}$
$e_{2} = \dfrac{6}{7}$
$e_{3} = \dfrac{-6}{7}$
Applying Newton's Identities again,
$P_{1'} = e_{1} = \dfrac{-9}{7}$
$P_{2'} = e_{1}P_{1'} - 2e_{2} = \dfrac{81}{49} -2 \times \dfrac{6}{7} = \dfrac{-3}{49}$
$P_{3'} = e_{1}P_{2'} - e_{2}P_{1'} + 3e_{3} = \dfrac{27}{343} + \dfrac{54}{49} + \dfrac{18}{7}= \dfrac {27 + 378 + 882}{343}= \dfrac{1287}{343}$
$M = \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} = P_{3'} = \dfrac{1287}{343}$
Comparing we get, $p = 1287 , q = 343$
$p + q = 1287 + 343 = 1630$