If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\frac{1}{a}=p,\frac{1}{b}=q,\frac{1}{c}=r$
$p+q+r=1$
$pqr=2$
$$pq+qr+rp=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$$
$$=\frac{ab+bc+b^2}{(abc)^2}$$
$$=4\bigg(\frac{1}{pq}+\frac{1}{qr}+\frac{1}{q^2}\bigg)$$
$$pq+qr+rp=4\bigg(\frac{pq+qr+rp}{pq^2r}\bigg)$$
$$pq^2r=4$$
$$\implies q=2 \implies b=\frac{1}{2}$$
So p, r are roots of $x^2+x+1=0$
$\implies p^3=q^3=1$
But this condition gives a different value of the required expression, so what am I doing wrong? Please tell me the right solution.

Hint:
Set $x^3=y$
$$(2x^3-1)^3=-(x^2+x)^3$$
$$(2y-1)^3=-(y)^2-y-2y\iff?\ \ \ \ (2)$$  whose roots are $a^3,b^3,c^3$
Now $$z=\dfrac1{a^3}+\dfrac1{b^3}-\dfrac1{c^3}=\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}-\dfrac2{c^3}=\dfrac{a^3b^3+b^3c^3+c^3a^3}{(abc)^3}-\dfrac2{c^3}$$
$$\implies\dfrac2{c^3}=?\iff c^3=?$$
Now as $c^3$ is a root of $(2),$ replace the value of $c^3$ in terms of $z$ to form a cubic equation $z$
