<s>[INST] 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.
 [/INST] 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
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=-6\cdot 8\cdot 8 =-384$$
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