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.
 A: You have a few mistakes:

*

*$(b+\frac1y)(y+\frac1a)(a+\frac1b)\neq a^2b^2y^2+\frac1{aby}$, as you write above

*$(b+\frac1y)(y+\frac1a)(a+\frac1b)\neq \frac{a^2b^2y^2+1}{aby}$, as you apply the formula you wrote

The actual solution is:
$$(b+\frac1y)(y+\frac1a)(a+\frac1b)=\frac1{aby}(yb+1)(ay+1)(ab+1)=\frac1{aby}(aby^2+ay+yb+1)(ab+1)=\frac1{aby}(a^2b^2y^2+a^2by+ab^2y+aby^2+ab+ay+yb+1)=aby+a+b+y+\frac{ab+ay+yb}{aby}+\frac1{aby}$$
Now, using Vieta's Formulas, we have that $aby=\frac{1}{3},ab+ay+yb=0,a+b+y=-\frac{8}{3}$.  This gives $(b+\frac1y)(y+\frac1a)(a+\frac1b)=\frac{1}{3}-\frac{8}{3}+0+3=\frac23$
A: Write $(b+1/y)(y+1/a)(a+1/b)$ as $\frac{(by+1)(ya+1)(ab+1)}{yab}$ by finding a common denominator for each bracket. Expanding this gives:
$$\frac{(aby)^2+aby(a+b+y)+ab+by+ya+1}{aby}$$
A: 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,
$$P= -3c = \frac 23$$
A: Hint:
$$(b+\frac{1}{y})(y+\frac{1}{a})(a+\frac{1}{b})=aby + \frac{1}{aby} + (a + b + y) + (\frac{1}{a} + \frac{1}{b} + \frac{1}{y})$$
(Here is the WolframAlpha verification for this computation.)
Can you finish?
