<s>[INST] If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which is equal to
$a^2b^2y^2 + 1/aby$
Using Vieta's formula I get:
$aby = -d/a$
$aby = 1/3$
Subbing in original:
$(1/9 + 1)/1/3$
which is...
10/27
The answer is supposedly 2/3, want to know what I did wrong.
 [/INST] You can also calculate the expression directly using the fact that
$p(x) = 3(x-a)(x-b)(x-y)\Rightarrow 3aby = 1$.
To do so, note that
\begin{eqnarray*} P
& = & (b+1/y)(y+1/a)(a+1/b) \\
& = & \frac{(by+1)(ay+1)(ab+1)}{aby} \\
& \stackrel{aby=\frac 13}{=} & 3\left(\frac 1{3a}+1\right)\left(\frac 1{3b}+1\right)\left(\frac 1{3y}+1\right) \\
& \stackrel{aby=\frac 13}{=} & \frac 13(1+3a)(1+3b)(1+3y) 
\end{eqnarray*}
Now, a standard trick uses the observation
\begin{eqnarray*} p\left(\frac{t-1}3\right)
& = & 3\left(\frac{t-1}3-a\right)\left(\frac{t-1}3-b\right)\left(\frac{t-1}3-y\right) \\
& = & \frac 19\left(t-(1+3a)\right)\left(t-(1+3b)\right)\left(t-(1+3y)\right)
\end{eqnarray*}
So, you only need the constant member $$c =-\frac 19(1+3a)(1+3b)(1+3y)=-\frac 13 P$$ of
$$p\left(\frac{t-1}3\right)= 3\left(\frac{t-1}3\right)^3 + 8\left(\frac{t-1}3\right)^2 -1 $$$$\Rightarrow c= -\frac 19+\frac 89-1 = -\frac 29$$
Hence,
$$P= -3c = \frac 23$$
</s>