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.
 A: Although this solution is mentioned in the comments to the question, I think it should be put here as an answer because of its simplicity and naturalness.
Looking at the given equation $2x^3+x^2+x-1=0$ one would quickly check possible rational roots using the Rational Root Test: 
$$\text{To check: }\pm 1, \pm\frac 12$$
This leads to the root $a=\frac 12$. Factoring gives now
$$2x^3+x^2+x-1 = 2\left(x-\frac 12 \right)\underbrace{(x^2+x+1)}_{=\frac{x^3-1}{x-1}}$$
Hence, the other two roots are the complex conjugated 3rd roots of $1$:
$$b^3=c^3=1$$
Now, plugging in, we get
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=-6\cdot 8\cdot 8 =-384$$
A: I think the answer is not equal to $16$.
Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3u$, $\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=3v^2$ and $\frac{1}{abc}=w^3$.
Thus, $u=\frac{1}{3},$ $v^2=-\frac{1}{3}$ and $w^3=2$.
Thus, $$\prod_{cyc}\left(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\right)=\prod_{cyc}\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{2}{c^3}\right)=$$
$$=(27u^3-27uv^2+3w^3)^3-2(27u^3-27uv^2+3w^3)^3+$$
$$+4(27u^3-27uv^2+3w^3)(27v^6-27uv^2w^3+3w^6)-8w^9=-384.$$
A: 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 $z$
A: $$2x^3+x^2+x-1=0`~~~(1)$$
$a,b, c$ 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 $64$
