If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16.

Question

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.

Answer 1

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, $$\prod_{cyc}\left(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\right)=\prod_{cyc}\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{2}{c^3}\right)=$$ $$=(27u^3-27uv^2+3w^3)^3-2(27u^3-27uv^2+3w^3)^3+$$ $$+4(27u^3-27uv^2+3w^3)(27v^6-27uv^2w^3+3w^6)-8w^9=-384.$$

Answer 2

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 $z$

Answer 3

$$2x^3+x^2+x-1=0`~~~(1)$$ $a,b, c$ 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 $64$

Answer 4

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

$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=-6\cdot 8\cdot 8 =-384$$