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Let $x^3+ax^2+bx+c=0$ are $\alpha, \beta, \gamma$. Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$

My attempt,

As I know from the original equation,

$\alpha+\beta+\gamma=-a$

$\alpha\beta+\beta\gamma+\alpha\gamma=b$

$\alpha\beta\gamma=-c$

I've tried to expand $(\alpha+\beta+\gamma)^3$ which is equal to $\alpha^3+\beta^3+\gamma^3+3\alpha^2\beta+3\alpha\beta^2+3\alpha^2\gamma+6\alpha\beta\gamma+3\beta^2\gamma+3\gamma^2\alpha+3\gamma\beta$

Basically, I know I've to find what's the value of $\alpha^3+\beta^3+\gamma^3$, $\alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3$ and $\alpha^3\beta^3\gamma^3$. But I m stuck at it. I would appreciate can someone explain and guide me to it. Thanks a lot.

By the way, I would appreciate if someone provides another tactics to solve this kind of routine question. Thanks a lot.

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  • $\begingroup$ Use the fact that $x^3 +ax^2 + bx + c = (x - \alpha)(x - \beta)(x - \gamma)$. $\endgroup$ – Hans Engler May 17 '17 at 14:47
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Let $p_k = \alpha^k + \beta^k + \gamma^k$. By Newton identities, we have

$$\begin{align} p_1 + a = 0 &\implies p_1 = -a\\ p_2 + ap_1 + 2b = 0&\implies p_2 = a^2 - 2b\\ p_3 + ap_2 + bp_1 + 3c = 0&\implies p_3 = -a^3 + 3ab - 3c \end{align} $$ In particular, we have $$\alpha^3 + \beta^3 + \gamma^3 = p_3 = -a^3 + 3ab - 3c$$

Apply Newton identities to the polynomial

$$x^3 + \frac{b}{c} x^2 + \frac{a}{c} x + \frac{1}{c}$$ which have roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$, we get

$$\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} = -\frac{b^3}{c^3} + 3\frac{ba}{c^2} - \frac{3}{c}\tag{*1}$$

By Vieta's formula, we have $\alpha\beta\gamma = - c$. Multiply LHS of $(*1)$ by $\alpha^3\beta^3\gamma^3$ and RHS by the same number $-c^3$, we get $$\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3 = 3c^2 - 3abc + b^3$$

As a result, the polynomial with roots $\alpha^3, \beta^3, \gamma^3$ is ( by Vieta's formula again)

$$x^3 - (\alpha^3+\beta^3+\gamma^3) x^2 + (\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3) - \alpha^3\beta^3\gamma^3\\ = x^3 + (a^3 -3ab + 3c)x^2 + (3c^2 - 3abc + b^3) x + c^3 $$

How to remember the Newton identifies

In general, given a polynomial of the form $$P(x) = x^n + a_1 x^{n-1} + \cdots + a_{n-1} x + a_n$$ with roots $\lambda_1, \ldots, \lambda_n$. The sequence of numbers $p_k = \sum_{i=1}^n\lambda_i^n$ satisfies a bunch of identities. $$\begin{cases} p_k + a_1 p_{k-1} + a_2 p_{k-2} + \cdots + a_{k-1} p_1 + \color{red}{k} a_k = 0, & k \le n\\ p_k + a_1 p_{k-1} + a_2 p_{k-2} + \cdots + a_{n-1} p_{k-n+1} + a_n p_{k-n} = 0, & k > n \end{cases}$$

For any particular $k$, you can obtain the corresponding identity by multiplying $p_{k-\ell}$ term with $a_\ell$ and sum over all available $0 \le \ell \le n$. When $k \le n$, you will have terms like $p_0$, $p_{-1}$,... Just replace all appearance of $p_0$ by $\color{red}{k}$ and forget all $p_\ell$ with negative $\ell$.

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  • $\begingroup$ You're really good at algebra! Are you a teacher in Hong Kong? $\endgroup$ – Mathxx May 17 '17 at 15:24
  • $\begingroup$ @Mathxx Nope, I'm just an amateur. $\endgroup$ – achille hui May 17 '17 at 15:26
  • $\begingroup$ How to get this ? $x^3 + \frac{b}{c} x^2 + \frac{a}{c} x + \frac{1}{c}$ $\endgroup$ – Mathxx May 17 '17 at 15:42
  • $\begingroup$ @Mathxx Given a polynomial $f(x)$ with degree $n$. If its roots are located at $\lambda_1,\lambda_2,\ldots,\lambda_n$, then $x^n f(\frac1x)$ has root located at $\lambda_1^{-1},\ldots,\lambda_n^{-1}$. In short, you obtain this polynomial with reciprocal roots by reversing the ordering of the coefficients. $\endgroup$ – achille hui May 17 '17 at 15:55
  • $\begingroup$ I don't know how to thank you. Your solution is very clear and I've understood fully. I'll redo the question myself hopefully. $\endgroup$ – Mathxx May 17 '17 at 15:58
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Hint

$$\alpha^3+a\alpha^2+b\alpha+c=0$$

$$\beta^3+a\beta^2+b\beta+c=0$$

$$\gamma^3+a\gamma^2+b\gamma+c=0$$

sum every equation:

$$(\alpha^3+\beta^3+\gamma^3)+a(\alpha^2+\beta^2+\gamma^2)+b(\alpha+\beta+\gamma)+3c=0$$

so you just need to find $\alpha^2+\beta^2+\gamma^2$. But

$$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha \gamma+\beta \gamma)$$

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  • $\begingroup$ It's really a brilliant way.! $\endgroup$ – Mathxx May 17 '17 at 14:48
  • $\begingroup$ Where is $(\alpha + \beta + \gamma)^3$?? $\endgroup$ – Hans Engler May 17 '17 at 14:48
  • $\begingroup$ @Arnaldo But how do I find $\alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3$ $\endgroup$ – Mathxx May 17 '17 at 15:08
  • $\begingroup$ @Mathxx: Sorry I didn't reply! I was busy! $\endgroup$ – Arnaldo May 17 '17 at 15:57
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If we replace $x$ by $x^{1/3}$, we obtain $x + ax^{2/3} + bx^{1/3} + c = 0$. This equation does have the required roots, but it's not a polynomial.

To get the good ol' cubic form, write it as $x + c = -(ax^{2/3} + bx^{1/3})$ and cube both sides so that$$(x + c)^3 = -(ax^{2/3} + bx^{1/3})^3 = -(a^3 x^2 + b^3 x + 3abx(ax^{2/3} + bx^{1/3})) = -(a^3 x^2 + b^3 x - 3abx(x+c))$$

This method wouldn't generate extraneous roots because we're not squaring anything here.

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