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.

$$2x^3+x^2+x-1=0`~~~(1)$$
$a,b, c$ are roots of the cubic, let $$s=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{a^3b^3+b^3c^3+c^3a^3}{(abc)^3}=0$$ Let one root od the new cubic be $$y=-\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=s-\frac{2}{a^3}=s-\frac{2}{x^3} \implies x=(\frac{2}{-y})^{1/3}~~~~(2)$$
From (1) we can write:
$$(2x^3+1)^3=(x^2+x)^3 \implies (2x^3+1)^3=x^6+x^3+3x^3(x^2+x)$$
$$\implies (2x^3+1)^3-[x^6+x^3+3x^3(-3x^3-1)] \implies 8x^9++17x^6+8x^3+1=0~~~(3)$$ 
Putting this transformation (2) in (3), we a cubic in $y$ as
$$y^3-16y^2+68y-64=0~~~~(4)$$ The required expression is nothing but $y_1 y_2 y_3$ and its value from (4) is $64$
