Cubic problem regarding solutions to a cubic.

Question

Let $x_1$, $x_2$, $x_3$ be the solutions of the equation $x^3 - 3x^2 + x - 1 = 0$. Determine the values of: $$\frac{1}{x_1x_2}+\frac{1}{x_2x_3}+\frac{1}{x_3x_1}$$ and $$x_1^3 + x_2^3 + x_3^3$$

Thanks for any help.

Answer 1

We use the usual relations between the roots and the coefficients of the polynomial (obtained by expanding $(x-x_1)(x-x_2)(x-x_3)$ and comparing coefficients).

(1). We have $x_1+x_2+x_3=3,x_1x_2x_3=1$. Dividing gives $\frac{1}{x_1x_2}+\frac{1}{x_2x_3}+\frac{1}{x_3x_1}=3$.

(2). We also have $x_1x_2+x_2x_3+x_3x_1=1$, so $x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3$ $=27-9+3=21$.

Answer 2

Hint: If $x_1, x_2, x_3$ are roots of the equation then $x^3 - 3x^2 + x - 1$, can you determine $x_1 + x_2 + x_3$, $x_1x_2 + x_2x_3 + x_3x_1$, and $x_1x_2x_3$?

EDIT: As per a commenter, I'll be a little less vague.

Once you compute the above values, then you can use the following:

Let $S_1 = x_1 + x_2 + x_3$, $S_2 = x_1x_2 + x_2x_3 + x_3x_1$, $S_3 = x_1x_2x_3$.

\begin{align*} x_1^3 + x_2^3 + x_3^3 &= (x_1 + x_2 + x_3)^3 - 3(x_1^2x_2 + x_1^2x_3 + x_2^2x_1 + x_2^2x_3 + x_3^2x_1 + x_3^2x_2) - 6x_1x_2x_3 \\ &= S_1^3 - 3x_1(x_1x_2 + x_1x_3) - 3x_2(x_2x_1 + x_2x_3) - 3x_3(x_3x_1 + x_3x_2) - 6S_3 \\ &= S_1^3 - 3x_1(S_2 - x_2x_3) + 3x_2(S_2 - x_3x_1) - 3x_3(S_2 - x_1x_2) - 6S_3$$ \\ &= S_1^3 - 3(x_1 + x_2 + x_3)S_2 - 3x_1x_2x_3 - 2x_2 x_3x_1 -3x_3 x_1 x_2 - 6S_3 \\ &= S_1^3 - 3S_1 S_2 - 15S_3. \end{align*}

And \begin{align*} \frac{1}{x_1x_2} + \frac{1}{x_2x_3} + \frac{1}{x_3x_1} &= \frac{S_1}{S_3}. \end{align*}

Answer 3

Given $$ x^3-3x^2+x-1=(x-x_1)(x-x_2)(x-x_3) \tag{1}$$ we have: $$ \left\{\begin{array}{ccccc}e_1 &=& x_1+x_2+x_3 &=& 3\\ e_2 &=& x_1 x_2+x_2 x_3+x_1 x_3 &=& 1\\ e_3 &=& x_1 x_2 x_3 &=& 1 \\ \end{array}\right.\tag{2}$$ by just expanding the RHS of $(1)$. Moreover, any $x_i$ fullfills $$ x_i^3 = 3x_i^2 - x_i + 1 \tag{3}$$ hence by $(1)$

$$ \frac{1}{x_1 x_2}+\frac{1}{x_2 x_3}+\frac{1}{x_1 x_3} = \frac{x_1+x_2+x_3}{x_2 x_2 x_3}=\frac{e_1}{e_3} = \color{red}{3} \tag{4}$$

and by $(3)$

$$ \begin{eqnarray*}p_3 = x_1^3+x_2^3+x_3^3 &=& 3\sum_{k=1}^{3} x_i^2 - e_1+3\\ &=& 3\left(e_1^2-2e_2\right)\\&=&3\cdot 7 = \color{red}{21}.\tag{5}\end{eqnarray*} $$