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using vietas transformation solve $x^3-6x+7$ in exact form.

The factor theorem won't work, as it doesn't factorize. It's known as a depressed cubic equation. Please help. EDIT: Sorry I'm new to this. Putting the equation into DESMOS the only value for $x$ is $-2.901$ (3dp).

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    What have you tried? Did you consider their hint? If yes, what are you stuck at? – Calvin Lin Jun 19 '23 at 21:39
  • To avoid closing, show some of your effort at solving. – Piita Jun 19 '23 at 21:42
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    The hint makes no sense to me either, but if you change the problem to one of solving $x^3+6x+7$, it becomes much easier to solve (although the hint still seems overkill). Think about how you could get $-7$ as the sum of two cubes (where you are allowed to take the cube of a negative number). I suspect the negative sign before $6x$ is an error in the problem statement. – David K Jun 19 '23 at 23:36
  • @DavidK There is definitely no error in the problem statement. Do you know any ideas or methods where I can use the hint? –  Jun 20 '23 at 00:15
  • @CalvinLin I've edited it with some of my recent working, but I still don't know how to apply the hint. Any ideas? –  Jun 20 '23 at 00:16
  • By "error in the problem statement" I merely meant that it does not work the way it was intended to work. I gather that you have checked your transcription and it is accurate. Therefore the error (if there is one) was made before the problem was presented to you. It is also possible that the problem statement is exactly what is is meant to be and that the solution given by dxiv is the one intended. – David K Jun 20 '23 at 01:09

1 Answers1

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One way to make sense of the hint would be to look for $\,m,n\,$ such that:$$(m+n)^3-3mn(m+n)-(m^3+n^3) = (m+n)^3 - 6 (m+n) + 7$$

Identifying coefficients of $(m+n)$ gives $\,mn=2\,$ and $\,m^3+n^3=-7\,$. Eliminating $\,n = \dfrac{2}{m}\,$ between the two equations: $$ m^3 + \frac{8}{m^3} = -7 \;\;\iff\;\; m^6 + 7 m^3 + 8 = 0 $$

The latter is a quadratic in $\,m^3\,$, so: $$ m^3 = \frac{-7 \pm \sqrt{17}}{2} \;\;\implies\;\; n^3 = \frac{8}{m^3} = \frac{-7 \mp \sqrt{17}}{2} $$

Then the real root of the original cubic is: $$ x = m + n = \sqrt[3]{\frac{-7 + \sqrt{17}}{2}} + \sqrt[3]{\frac{-7 - \sqrt{17}}{2}} $$

dxiv
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  • Here is a step-by-step of how this works for depressed cubics in the general case. – dxiv Jun 20 '23 at 00:35
  • +1. 1) I would think it might be worth pointing out for the OP that this is one possible general way of deriving the general formula of the cubic equation. e.g. $x^3+px+q=0$ and comparing $(a+b)^3-3ab(a+b)-(a^3+b^3)=0$ assuming that root is $x=a+b$ and we have $2$ complex solutions, since $m^3=a$ has $3$ roots . 2) Maybe OP might want how to $\frac {8}{m^3}= \frac{-7 \pm \sqrt{17}}{2} $, one way would be $$m^3\cdot n^3 =8\ m^3+n^3=-7$$ which implies, $m^3$ and $n^3$ are the roots of $u^2+7u+8=0$ which yields $u_1=m^3$ and $u_2=n^3$ . – lone student Jun 20 '23 at 00:46
  • Ahh, after that comment, I saw your comment . – lone student Jun 20 '23 at 00:48
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    @lonestudent Indeed, that's covered in the answer linked in my first comment. – dxiv Jun 20 '23 at 00:58