rzfansubs87

Answered

2022-07-01

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

Answer & Explanation

Sophia Mcdowell

Expert

2022-07-02Added 14 answers

You have a few mistakes:

$(b+\frac{1}{y})(y+\frac{1}{a})(a+\frac{1}{b})\ne {a}^{2}{b}^{2}{y}^{2}+\frac{1}{aby}$, as you write above

$(b+\frac{1}{y})(y+\frac{1}{a})(a+\frac{1}{b})\ne \frac{{a}^{2}{b}^{2}{y}^{2}+1}{aby}$, as you apply the formula you wrote

The actual solution is:

$(b+\frac{1}{y})(y+\frac{1}{a})(a+\frac{1}{b})=\frac{1}{aby}(yb+1)(ay+1)(ab+1)=\frac{1}{aby}(ab{y}^{2}+ay+yb+1)(ab+1)=\frac{1}{aby}({a}^{2}{b}^{2}{y}^{2}+{a}^{2}by+a{b}^{2}y+ab{y}^{2}+ab+ay+yb+1)=aby+a+b+y+\frac{ab+ay+yb}{aby}+\frac{1}{aby}$

Now, using Vieta's Formulas, we have that $aby=\frac{1}{3},ab+ay+yb=0,a+b+y=-\frac{8}{3}$. This gives $(b+\frac{1}{y})(y+\frac{1}{a})(a+\frac{1}{b})=\frac{1}{3}-\frac{8}{3}+0+3=\frac{2}{3}$

Desirae Washington

Expert

2022-07-03Added 5 answers

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$\begin{array}{rcl}P& =& (b+1/y)(y+1/a)(a+1/b)\\ & =& \frac{(by+1)(ay+1)(ab+1)}{aby}\\ & \stackrel{aby=\frac{1}{3}}{=}& 3(\frac{1}{3a}+1)(\frac{1}{3b}+1)(\frac{1}{3y}+1)\\ & \stackrel{aby=\frac{1}{3}}{=}& \frac{1}{3}(1+3a)(1+3b)(1+3y)\end{array}$

Now, a standard trick uses the observation

$\begin{array}{rcl}p\left(\frac{t-1}{3}\right)& =& 3(\frac{t-1}{3}-a)(\frac{t-1}{3}-b)(\frac{t-1}{3}-y)\\ & =& \frac{1}{9}(t-(1+3a))(t-(1+3b))(t-(1+3y))\end{array}$

So, you only need the constant member

$c=-\frac{1}{9}(1+3a)(1+3b)(1+3y)=-\frac{1}{3}P$

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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