For $3x^3-5x^2+12x-18=0$ find the value of $(1+\frac{x_1}{x_2})(1+\frac{x_2}{x_3})(1+\frac{x_3}{x_1})$ using Vieta's formulas

Question

For $3x^3-5x^2+12x-18=0$ find the value of $(1+\frac{x_1}{x_2})(1+\frac{x_2}{x_3})(1+\frac{x_3}{x_1})$ using Vieta's formulas.

I transformed it into $\dfrac{(x_2+x_1)(x_3+x_2)(x_1+x_3)}{x_1x_2x_3}$ , now I know the denominator equals $6$ but I'm stuck on the numerator.

Answer 1

Hint: \begin{eqnarray*} \sum \alpha^2 \beta =(\alpha+\beta+\gamma)( \alpha \beta+\beta \gamma+ \gamma \alpha)-3 \alpha \beta \gamma. \end{eqnarray*}

Answer 2

Hint.

Vieta's Theorem for cubic equations says: a cubic equation $x^3 + px^2 + qx + r = 0$ has three different roots $x_1, x_2, x_3$:

$$\begin{eqnarray*} -p &=& x_1 + x_2 + x_3 \\ q &=& x_1x_2 + x_1x_3 + x_2x_3 \\ -r &=& x_1x_2x_3 \end{eqnarray*}$$

Answer 3

use that $$-5=x_1-x_2-x_3$$ $$12=x_1x_2+x_1x_3+x_2x_3$$ $$-18=-x_1x_2x_3$$

Answer 4

Hint:   let $s=x_1+x_2+x_3=\dfrac{5}{3}$, then (with $P(x)$ being the original cubic):

$$(x_2+x_1)(x_3+x_2)(x_1+x_3)=(s-x_1)(s-x_2)(s-x_3)=\frac{1}{3}P(s)=\frac{1}{3}P\left(\frac{5}{3}\right)$$

Answer 5

One has

$$\begin{align} (x_1+x_2)(x_2+x_3)(x_1+x_3) &=(x_1x_2+x_1x_3+x_2^2+x_2x_3)(x_1+x_3)\\ &=x_1^2x_2+x_1x_2x_3+x_1^2x_3+x_1x_3^2+x_1x_2^2+x_2^2x_3+x_1x_2x_3+x_2x_3^2\\ &=x_1x_2(x_1+x_2+x_3)+x_1x_3(x_1+x_2+x_3)+x_2x_3(x_1+x_2+x_3)-x_1x_2x_3\\ &=(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)-x_1x_2x_3\end{align}$$

Answer 6

Simplify the second and third terms of the equation first $(1+\frac{x_1}{x_2})(1+\frac{x_2}{x_3})(1+\frac{x_3}{x_1})$:

$$\biggl(1+\frac{x_1}{x_2}\biggr)\biggl(\frac{x_1x_3+x_1x_2+x_2x_3}{x_3x_1}+\frac{x_3}{x_1} \biggr)$$ $$\biggl(1+\frac{x_1}{x_2}\biggr)\biggl(\frac{\frac ca}{x_3x_1}+\frac{x_3}{x_1} \biggr)$$ Then distribute the first term with the second term, and knowing where $x_3x_2+x_3x_1=\frac ca-x_1x_2$: $$\biggl(\frac{x_2+x_1}{x_2}\biggr)\biggl(\frac{\frac ca}{x_3x_1}\biggr)+\biggl(\frac{x_2+x_1}{x_2}\biggr)\cdot\frac{x_3}{x_1} $$ $$\frac{\frac ca\cdot(x_2+x_1)}{x_1x_2x_3}+\frac{x_3x_2+x_3x_1}{x_2x_1}$$

$$\frac{c(x_2+x_1)}{a(x_1x_2x_3)}+\frac{\frac ca-x_1x_2}{x_1x_2}$$ Which can be simplified to: $$\frac{c(x_2+x_1)}{a(x_1x_2x_3)}+\frac{c}{a}\cdot\frac1{x_1x_2}-1$$ $$\frac ca\biggl(\frac{x_1x_2^2+x_1^2x_2+x_1x_2x_3}{x_1^2x_2^2x_3}\biggr)-1$$ $$\frac ca\biggl(\frac1{x_1x_3}+\frac1{x_2x_3}+\frac1{x_1x_2}\biggr)-1$$ $$\frac ca\biggl(\frac{x_1+x_2+x_3}{x_1x_2x_3}\biggr)-1$$ $$\frac ca\cdot\biggl(\frac{-b}c\biggr)-1=\frac{-b}a-1$$ $$\therefore (1+\frac{x_1}{x_2})(1+\frac{x_2}{x_3})(1+\frac{x_3}{x_1})=\frac53-1=\frac23$$