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.

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 $(b+\frac1y)(y+\frac1a)(a+\frac1b)=\frac{1}{3}-\frac{8}{3}+0+3=\frac23$
