rzfansubs87

2022-07-01

If a b and y are the roots of $3{x}^{3}+8{x}^{2}-1=0$ find $\left(b+1/y\right)\left(y+1/a\right)\left(a+1/b\right)$

Sophia Mcdowell

You have a few mistakes:
$\left(b+\frac{1}{y}\right)\left(y+\frac{1}{a}\right)\left(a+\frac{1}{b}\right)\ne {a}^{2}{b}^{2}{y}^{2}+\frac{1}{aby}$, as you write above
$\left(b+\frac{1}{y}\right)\left(y+\frac{1}{a}\right)\left(a+\frac{1}{b}\right)\ne \frac{{a}^{2}{b}^{2}{y}^{2}+1}{aby}$, as you apply the formula you wrote
The actual solution is:
$\left(b+\frac{1}{y}\right)\left(y+\frac{1}{a}\right)\left(a+\frac{1}{b}\right)=\frac{1}{aby}\left(yb+1\right)\left(ay+1\right)\left(ab+1\right)=\frac{1}{aby}\left(ab{y}^{2}+ay+yb+1\right)\left(ab+1\right)=\frac{1}{aby}\left({a}^{2}{b}^{2}{y}^{2}+{a}^{2}by+a{b}^{2}y+ab{y}^{2}+ab+ay+yb+1\right)=aby+a+b+y+\frac{ab+ay+yb}{aby}+\frac{1}{aby}$
Now, using Vieta's Formulas, we have that $aby=\frac{1}{3},ab+ay+yb=0,a+b+y=-\frac{8}{3}$. This gives $\left(b+\frac{1}{y}\right)\left(y+\frac{1}{a}\right)\left(a+\frac{1}{b}\right)=\frac{1}{3}-\frac{8}{3}+0+3=\frac{2}{3}$

Desirae Washington

You can also calculate the expression directly using the fact that
$p\left(x\right)=3\left(x-a\right)\left(x-b\right)\left(x-y\right)⇒3aby=1$
To do so, note that$\begin{array}{rcl}P& =& \left(b+1/y\right)\left(y+1/a\right)\left(a+1/b\right)\\ & =& \frac{\left(by+1\right)\left(ay+1\right)\left(ab+1\right)}{aby}\\ & \stackrel{aby=\frac{1}{3}}{=}& 3\left(\frac{1}{3a}+1\right)\left(\frac{1}{3b}+1\right)\left(\frac{1}{3y}+1\right)\\ & \stackrel{aby=\frac{1}{3}}{=}& \frac{1}{3}\left(1+3a\right)\left(1+3b\right)\left(1+3y\right)\end{array}$
Now, a standard trick uses the observation
$\begin{array}{rcl}p\left(\frac{t-1}{3}\right)& =& 3\left(\frac{t-1}{3}-a\right)\left(\frac{t-1}{3}-b\right)\left(\frac{t-1}{3}-y\right)\\ & =& \frac{1}{9}\left(t-\left(1+3a\right)\right)\left(t-\left(1+3b\right)\right)\left(t-\left(1+3y\right)\right)\end{array}$
So, you only need the constant member
$c=-\frac{1}{9}\left(1+3a\right)\left(1+3b\right)\left(1+3y\right)=-\frac{1}{3}P$
of
$p\left(\frac{t-1}{3}\right)=3{\left(\frac{t-1}{3}\right)}^{3}+8{\left(\frac{t-1}{3}\right)}^{2}-1$
Hence,$P=-3c=\frac{2}{3}$

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